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AN  ELEMENTARY  TREATISE 

ON 

ELECTRIC  POWER  AND  LIGHTING 

PREPARED  FOR  STUDENTS  OF 

THE  INTERNATIONAL  CORRESPONDENCE  SCHOOLS 
SCRANTON,  PA. 


Volume 


ARITHMETIC 

MENSURATION 

MECHANICS 

WITH  PRACTICAL  "QUESTIONS  AND  EXAMPLES 


First  Edition 


SCRANTON 

THE  COLLIERY  ENGINEER  CO. 
1897 


235 


Entered  according  to  the  Act  of  Congress,  in  the  year  1897,  by  THE  COLLIERY 

ENGINEER  COMPANY,  in  the  office  of  the  Librarian  of  Congress. 

at  Washington. 


RURR  PRINTING  HOUSE, 
FRANKFORT  AND  JACOB  STREETS, 

NEW  YORK. 


SMI 
Utt 


PREFACE. 


The  Instruction  and  Question  Papers  which  are  furnished 
to  the  students  of  The  International  Correspondence 
Schools  become  so  badly  worn  and  soiled  that,  when  a  student 
has  completed  his  Course,  he  has  worn  out  his  Instruction 
Papers,  and  they  are  no  longer  suitable  for  reference  or  re- 
view. Since  the  Instruction  Papers  are  very  valuable,  es- 
pecially to  those  who  have  studied  them,  there  has  grown 
up  a  demand  for  Sets  of  the  Instruction  and  Question 
Papers,  indexed  for  convenient  reference,  and  durably  bound 
for  preservation.  Again,  many  of  our  students  can  spare 
but  little  time  for  study,  and  are,  therefore,  a  long  while 
passing  through  their  Courses.  These  students  also  desire 
the  Papers  in  bound  volumes  to  use  for  reference.  Other 
students  begin  Courses,  but  for-  various  reasons  are  unable 
to  complete  them,  and  feel  that,  having  paid  for  their  Scholar- 
ships, they  ought  to  have  the  text-books,  even  though  they 
can  not  finish  their  Courses. 

For  these  reasons,  we  have  decided  to  publish  all  of  the 
Instruction  and  Question  Papers  of  our  different  Technical 
Courses  in  volumes  bound  in  Half  Leather,  to  make  a  small 
advance  in  our  prices,  and  to  furnish  a  set  to  each  student  as 
soon  as  his  Scholarship  is  paid  for,  whether  he  has  completed 
his  studies  or  not. 

The  volumes  for  the  present  Course,  the  Electric  Power 
and  Lighting,  are  five  in  number: 

Volume  I  contains  the  Instruction  and  Question  Papers 
on  Arithmetic,  Mensuration,  and  Mechanics. 

Volume  II  contains  the  Instruction  and  Question  Papers 
on  Dynamos  and  Motors,  Electric  Lighting,  and  Electric 
Railways. 


iv  PREFACE. 

Volume  III  contains  the  Plates  and  the  instructions  for 
drawing  them.  It  forms  a  very  complete  Course  in  Me- 
chanical Drawing. 

Volume  IV  contains  the  Tables,  Rules,  and  Formulas  in 
common  use.  The  student  who  has  finished  his  Course  will 
find  these  Tables  of  great  service.  All  the  principal  for- 
mulas, with  the  definitions  of  the  letters  used  in  them,  and 
the  principal  rules,  are  conveniently  arranged  for  reference, 
so  that  the  student  can  save  the  labor  and  time  of  hunting 
them  out  in  the  Instruction  Papers. 

Volume  V  contains  the  answers  to  the  questions  in  the 
Question  Papers. 

These  Instruction  Papers  are  written  from  a  practical 
standpoint,  and  contain  only  such  information  as  the  student 
requires  in  order  to  obtain  a  good  working  knowledge  of 
the  subjects  which  form  the  groundwork  of  a  course  in 
Electric  Power  and  Lighting.  There  is  no  padding,  and 
the  student  gets  in  a  clear  and  concise  form  the  exact 
information  which  he  desires. 

To  keep  the  student  always  interested  in  his  work,  we  do 
not  give  him  half  a  dozen  ways  of  doing  the  same  thing; 
neither  do  we  enter  into  speculative  discussions  of  different 
subjects,  all  of  which  tend  to  confuse  the  student  and  leave 
him  in  doubt  as  to  which  one  (or  any  one)  is  correct ;  we 
give  him  the  best  or  most  suitable  rule,  formula,  or  method 
that  we  know  of,  without  mentioning  any  of  the  others.  In 
short,  the  Papers  are  written  for  practical  men,  and  all  the 
difficulties  encountered  by  a  student  studying  by  himself, 
particularly  those  which  are  due  to  a  definition  or  explanation 
being  too  technical,  abstract,  or  not  clear  enough  to  be 
readily  understood,  have  been  carefully  considered  and 
overcome. 

THE  INTERNATIONAL  CORRESPONDENCE  SCHOOLS. 


CONTENTS. 

ARITHMETIC.  PAGE 

Definitions,              1 

Notation  and  Numeration,     -----  i 

Addition,                   4 

Subtraction,    -  9 

Multiplication, 12 

Division,                                                        ...  17 
Cancelation,    --.-                  ---21 

Fractions,        -         -         -               .  -         -         -         -  23 

Decimals, 38 

Percentage,     -                            55 

Denominate  Numbers,   -  61 

Involution,      --------  7^3 

Evolution, -"  79 

Ratio, 96 

Proportion,  100 

MENSURATION    AND   USE   OF    LETTERS   IN    ALGEBRAIC 

.  *  FORMULAS. 

Formulas, 115 

Lines  and  Angles, 119 

Quadrilaterals,        -         -                 -        -        -         -  123 

The  Triangle, 120 

Polygons, 130 

The  Circle,                133 

The  Prism  and  Cylinder, 138 

The  Pyramid  and  Cone,                •  -         -         -         -  142 

The  Frustum  of  a  Pyramid  or  Cone,      ...  143 

The  Sphere  and  Cylindrical  Ring,          -        -        -  145 

MECHANICS. 

Matter  and  its  Properties, 149 

Motion  and  Velocity,       ------  153 

Force, -  156 


vi  CONTENTS. 

MECHANICS — continued.  PAGE 

Center  of  Gravity, 160 

Simple  Machines,    -------  165 

Pulleys,  -                  -       .  -    " 170 

Gear- Wheels, 175 

Fixed  and  Movable  Pulleys, 183 

The  Inclined  Plane,                  185 

The  Screw,      -         -         -         -         -         -         -         -  187 

Friction,                    -         -         -  .      -         -         -         -  189 

Centrifugal  Force, 194 

Specific  Gravity, 196 

Work  and  Energy,          -  '      -         -         -         -         -  197 

Belts,                                 " 201 

Horsepower  of  Gears,     ------  204 

Hydrostatics, 207 

Buoyant  Effects  of  Water, 219 

Pneumatics, 222 

Pneumatic  Machines,      -         -         -         -         -  231 

Pumps,    -                            ------  241 

Strength  of  Materials, 247 

Tensile  Strength  of  Materials,        -         -         -         -  248 

Chains,    -                  -  251 

Hemp  Ropes,                    -        -        -                  -         -  252. 

Wire  Ropes, *         -         -  253 

Crushing  Strength  of  Materials,     -         -         -         -  255 

Transverse  Strength  of  Materials,          "        -         -  261 

Shearing  Strength  of  Materials,     -               .   -.       -  266 

Line  Shafting,         -         -     •   -         -         ....  268 

QUESTIONS  AND  EXAMPLES. 

Arithmetic,     -         -         -         -         -         -         -         -  275 

Arithmetic,  continued,    -         -         -         -         -         -  283 

Mensuration    and    Use    of  'Letters   in  Algebraic 

Formulas,    -         -         -         ...         -         -  291 

Mechanics,       -         -         - 299 

Mechanics,  continued,     -         -         -         -         -     .     -  309 


ARITHMETIC 


DEFINITIONS. 

1.  Arithmetic  is  the  art  of  reckoning,  or  the  study  of 
numbers. 

2.  A  unit  is  one,  or  a  single  thing,  as  one,  one  bolt,  one 
pulley,  one  dozen. 

3.  A  number  is  a  unit  or  a  collection  of  units,  as  one, 
three  engines,  five  boilers. 

4.  The  unit  of  a  number  is  one  of  the  collection  of 
units   which   constitutes  the  number.     Thus,    the    unit    of 
twelve  is  one,  of  twenty  dollars  is  one  dollar,  of  one  hundred 
bolts  is  one  bolt. 

5.  A  concrete  number  is  a  number  applied  to  some 
particular  kind  of  object  or  quantity,  as  three  grate  bars, 
five  dollars,  ten  pounds. 

6.  An  abstract  number  is  a  number  that  is  not  ap- 
plied to  any  object  or  quantity,  as  three,  five,  ten. 

7.  Like  numbers  are  numbers  which  express  units  of 
the  same  kind,  as  6  days  and  10  days,  2  feet  and  5  feet. 

8.  Unlike  numbers  are  numbers  which  express  units 
of  different  kinds,  as   ten   months  and   eight    miles,  seven 
wrenches  and  five  bolts.      

NOTATION   AND  NUMERATION. 

9.  Numbers  are  expressed  in  three  ways:  (1)  by  words; 
(2)  by  figures;  (3)  by  letters. 

10.  Notation  is  the  art  of  expressing  numbers  by  fig- 
ures or  letters. 

11.  Numeration  is  the  art  of  reading  the   numbers 
which  have  been  expressed  by  figures  or  letters. 


2  ARITHMETIC. 

1 2.  The  Arabic  notation  is  the  method  of  expressing 
numbers   by  figures.     This   method  employs  ten  different 
figures  to  represent  numbers,  viz. : 

Figures         0          123456789 

naught,    one      two     three    four     five       six     seven   eight   nine 
Names        cipher 
or  zero; 

The  first  character  (0)  is  called  naught,  cipher,  or  zero, 

and,  when  standing  alone,  has  no  value. 

The  other  nine  figures  are  called  digits,  and  each  one 
has  a  value  of  its  own. 

Any  whole  number  is  called  an  integer. 

13.  As  there  are  only  ten  figures  used   in  expressing 
numbers,  each  figure  must  express  a  different  value  at  differ- 
ent times. 

14.  The  value  of  a  figure  depends  upon  its  position  in 
relation  to  others. 

15.  Figures  have  simple  values  and  local  or  relative 
values. 

16.  The  simple  value  of  a  figure  is  the  value  it  ex- 
presses when  standing  alone. 

17.  The  local  or  relative  value  is  the  increased  value 
it  expresses  by  having  other  figures  placed  on  its  right. 

For  instance,  if  we  see  the  figure  6  standing  alone, 

thus 6 

we  consider  it  as  six  units,  or  simply  six. 

Place  another  6  to  the  left  of  it;  thus 66 

The  original  figure  is  still  six  units,  but  the  second 

one  is  ten  times  6,  or  6  tens. 

If  a  third  6  be  now  placed  still  one  place  further  to 

the  left,  it  is  increased  in  value  ten  times  more,  thus 

making  it  6  hundreds 666 

A  fourth  6  would  be  6  thousands 6666 

A  fifth  6  would  be  6  tens  of  thousands,  or  sixty 

thousand 66666 

A  sixth  6  would  be  6  hundreds  of  thousands  .     666666 

A  seventh  6  would  be  6  millions .  ,   6666666 

The  entire  line  of  seven  figures  is  read  six  millions,  six 

hundred  sixty-six  tJwusands,  six  hundred  sixty-six. 


ARITHMETIC. 


18.  The  increased  value  of  each  of  these  figures  is  its 
local  or  relative  value.      Each  figure  is  ten  times  greater  in 
value  than  the  one  immediately  on  its  right. 

19.  The  cipher  (0)  has  no  value  itself,  but  it  is  useful  in 
determining  the  place  of  other  figures.     To  represent  the 
number  four  hundred  five,  two  digits  only  are  necessary,  one 
to  represent  four  hundred,  and  the  other  to.  represent  five 
units ;  but  if  these  two  digits  are  placed  together,  as  45,  the 
4  (being  in  the  second  place)  will  mean  4  tens.     To  mean  4 
hundreds,  the  4  should  have  two  figures  on  its  right,  and  a 
cipher  is  therefore  inserted  in  the  place  usually  given  to  fens, 
to  show  that  the  number  is  composed  of  hundreds  and  units 
only,  and  that  there  are  no  tens.     Four  hundred  five  is  there- 
fore expressed  as  405.     If  the  number  were  four  thousand 
and  five,  two  ciphers  would  be  inserted;  thus,  4005.      If  it 
were  four  hundred  fifty,  it  would  have  the  cipher  at  the  right- 
hand  side  to  show  that  there  were  no  units,  and  only  hun- 
dreds and tens ;  thus,  450.     Four  thousand  and  fifty  would  be 
expressed  4050,  the  first  cipher  indicating  that  there  are  no 
hundreds  and  the  second  that  there  are  no  units. 

NOTE. — When  speaking  of  the  figures  of  a  number  by  referring  to 
them  as  first  figure,  second  figure,  etc.,  always  begin  to  count  at  the 
left.  Thus,  in  the  number  41,625,  4  is  the  first  figure,  6  the  third 
figure,  5  the  fifth  or  last  figure,  etc. 

20.  In  reading  figures,  it  is  usual  to  point  off  the  num- 
ber into  groups  of  three  figures  each,  beginning  with  the 
right-hand   or   units   column,  a  comma  (,)  being  used  to 
point  off  these  groups. 


Billions. 

Millions. 

Thousands. 

Units. 

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4  ARITHMETIC. 

In  pointing  off  these  figures,  begin  at  the  right-hand  figure 
and  count — units,  tens,  hundreds ;  the  next  group  of  three 
figures  is  thousands,  therefore,  we  insert  a  comma  (,)  before 
beginning  with  them.  Beginning  at  the  figure  5,  we  say 
thousands,  tens  of  thousands,  hundreds  of  thousands,  and  in- 
sert another  comma ;  we  next  read  millions,  tens  of  millions, 
hundreds  of  millions,  and  insert  another  comma;  we  then 
read  billions,  tens  of  billions,  hundreds  of  billions. 

The  entire  line  of  figures  would  be  read:  Four  hundred 
thirty-two  billions,  one  hundred  ninety-eight  millions,  seven 
hundred  sixty-five  thousands,  four  hundred  thirty-two.  When 
we  thus  read  a  line  of  figures  it  is  called  numeration,  and 
if  the  numeration  be  changed  back  to  figures,  it  is  called 
notation. 

For  instance,  the  writing  of  the  figures, 

72,584,623, 

would  be  the  notation,  and  the  numeration  would  be 
seventy-two  millions,  five  hundred  eighty-four  thousands,  six 
hundred  twenty-three. 

21 .  NOTE.— It  is  customary  to  leave  the  s  off  the  words  millions, 
thousands,  etc. ,  in  cases  like  the  above,  both  in  speaking  and  writing ; 
hence,  the  above  would  usually  be  expressed,  seventy-two  million,  five 
hundred  eighty-four  thousand,  six  hundred  twenty-three. 

22.  The  four  fundamental  processes  of  Arithmetic  are 
addition,  subtraction,  multiplication,  and  division. 

They  are  called  fundamental  processes,  because  all  opera- 
tions in  Arithmetic  are  based  upon  them. 


ADDITION. 

23.  Addition  is  the  process  of  finding  the  sum  of  two  or 
more  numbers.     The  sign  of  addition  is  -f  .     It  is  read  plus, 
and  means  more.     Thus,  5  -f-  6  is  read  5  plus  6,  and  means 
that  5  and  6  are  to  be  added. 

24.  The  sign  of  equality  is  =  .      It  is  read  equals  or  is 
equal  to.     Thus,  5  +  6  =  11  may  be  read  5  plus  6  equals  11. 

25.  Like  numbers  can  be  added,  but  unlike  numbers  can- 
not.    Thus,  6  dollars  can  be  added  to  7  dollars,  and  the  sum 
will  be  13  dollars,  but  6  dollars  cannot  be  added  to  ^  feet. 


ARITHMETIC. 


26.     The  following  table  gives  the  sum  of  any  two  num- 
bers from  1  to  12: 

TABLE    1. 


1  and    1  are    2 

2  and    1  are    3 

3  and    1  are    4 

4  and    1  are    5 

1  and    2  are    3 

2  and    2  are    4 

3  and    2  are    5 

4  and    2  are    6 

1  and    3  are    4 

2  and    3  are    5 

3  and    3  are    6 

4  and    3  are    7 

1  and    4  are    5 

2  and    4  are    6 

3  and    4  are    7 

4  and    4  are    8 

1  and    5  are    6 

2  and    5  are    7 

3  and    5  are    8 

4  and    5  are    9 

1  and    6  are    7 

2  and    6  are    8 

3  and    6  are    9 

4  and    6  are  10 

1  and    7  are    8 

2  and    7  are    9 

3  and    7  are  10 

4  and    7  are  11 

1  and    8  are    9 

2  and    8  are  10 

3  and    8  are  11 

4  and    8  are  12 

1  and    9  are  10 

2  and    9  are  11 

3  and    9  are  12 

4  and    9  are  13 

1  and  10  are  11 

2  and  10  are  12 

3  and  10  are  13 

4  and  10  are  14 

1  and  11  are  12 

2  and  11  are  13 

3  and  11  are  14 

4  and  11  are  15 

1  and  12  are  13 

2  and  12  are  14 

3  and  12  are  15 

4  and  12  are  16 

5  and    1  are    6 

6  and    1  are    7 

7  and    1  are    8 

8  and    1  are    9 

5  and    2  are    7 

6  and    2  are    8 

7  and    2  are    9 

8  and    2  are  10 

5  and    3  are    8 

6  and    3  are    9 

7  and    3  are  10 

8  and    3  are  11 

5  and    4  are    9 

6  and    4  are  10 

7  and    4  are  11 

Sand    4  are  12 

5  and    5  are  10 

6  and    5  are  11 

7  and    5  are  12 

8  and    5  are  13 

-r>  and    6  are  11 

6  and    6  are  12 

7  and    6  are  13 

8  and    6  are  14 

5  and    7  are  12 

6  and    7  are  13 

7  and    7  are  14 

8  and    7  are  15 

5  and    8  are  13 

6  and    8  are  14 

7  and    8  are  15 

8  and    8  are  16 

5  and    9  are  14 

6  and    9  are  15 

7  and    9  are  16 

8  and    9  are  17 

5  and  10  are  15 

6  and  10  are  16 

7  and  10  are  17 

8  and  10  are  18 

5  and  11  are  16 

6  and  11  are  17 

7  and  11  are  18 

8  and  11  are  19 

5  and  12  are  17 

6  and  12  are  18 

7  and  12  are  19 

8  and  12  are  20 

9  and    1  are  10 

10  and    1  are  11 

11  and    1  are  12 

12  and    1  are  13 

9  and    2  are  11 

10  and    2  are  12 

11  and    2  are  13 

12  and    2  are  14 

9  and    3  are  12 

10  and    3  are  13 

11  and    3  are  14 

12  and    3  are  15 

9  and    4  are  13 

10  and    4  are  14 

11  and    4  are  15 

12  and    4  are  16 

9  and    5  are  14 

10  and    5  are  15 

11  and    5  are  16 

12  and    5  are  17 

9  and    6  are  15 

10  and    6  are  16 

11  and    6  are  17 

12  and    6  are  18 

9  and    7  are  16 

10  and    7  are  17 

11  and    7  are  18 

12  and    7  are  19 

9  and    8  are  17 

10  and    8  are  18 

11  and    8  are  19 

12  and    8  are  20 

9  and    9  are  18 

10  and    9  are  19 

11  and    9  are  20 

12  and    9  are  21 

9  and  10  are  19 

10  and  10  are  20 

11  and  10  are  21 

12  and  10  are  22 

9  and  11  are  20 

10  and  11  are  21 

11  and  11  are  22 

12  and  11  are  23 

9  and  12  are  21 

10  and  12  are  22 

11  and  12  are  23 

12  and  12  are  24 

This  table  should  be  carefully  committed  to  memory.  Since  0  has 
no  value,  the  sum  of  any  number  and  0  is  the  number  itself;  thus,  17 
and  0  are  17. 

27.  For  addition,  place  the  numbers  to  be  added  directly 
under  each  other,  taking  care  to  place  units  under  units,  tens 
under  tens,  liundreds  under  hundreds  ^  and  so  on. 

When  the  numbers  are  thus  written,  the  right-hand  figure 
of  one  number  is  placed  directly  under  the  right-hand  figure 


6  ARITHMETIC. 

of  the  number  above  it,  thus  bringing  the  unit  figures  of  all 
the  numbers  to  be  added  in  the  same  vertical  line.  Proceed 
as  in  the  following  examples: 

28.  EXAMPLE.— What  is  the  sum  of  181,  222,  21,  2,  and  413  ? 
SOLUTION. —  131 

222 

21 

2 

413 

sum    789    Ans. 

EXPLANATION. — After  placing  the  numbers  in  proper 
order,  begin  at  the  bottom  of  the  right-hand  or  units 
column,  and  add,  mentally  repeating  the  different  sums. 
Thus,  three  and  two  are  five  and  one  are  six  and  two  are 
eight  and  one  are  nine,  the  sum  of  the  numbers  in  units 
column.  Place  the  9  directly  beneath  as  the  first  or  units 
figure  in  the  sum. 

The  sum  of  the  numbers  in  the  next  or  tens  column  equals 
8  tens,  which  is  the  second  or  tens  figure  in  the  sum. 

The  sum  of  the  numbers  in  the  next  or  hundreds  column 
equals  7  hundreds^  which  is  the  third  or  hundreds  figure  in 
the  sum. 

The  sum  or  answer  is  789. 

29.  EXAMPLE.— What  is  the  sum  of  425,  36,  9,215,  4,  and  907  ? 
SOLUTION. —  425 

36 

9215 

4 

907 


27 

60 

1500 
9000 

sum  10587  Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  the  first  or 
units  column  is  seven  and  four  are  eleven  and  five  are  six- 
teen and  six  are  twenty-two  and  five  are  twenty-seven,  or 
37  units;  i.  e.,  two  tens  and  seven  units.  Write  27  as  shown. 


ARITHMETIC.  7 

The  sum  of  the  numbers  in  the  second  or  tens  column  is 
six  tens,  or  GO.  Write  60  underneath  27  as  shown.  The 
sum  of  the  numbers  in  the  third  or  hundreds  column  is  15 
hundreds,  or  1,500.  Write  1,500  under  the  two  preceding 
results  as  shown.  There  is  only  one  number  in  the  fourth 
or  thousands  column,  nine,  which  represents  9,000.  Write 
9,000  under  the  three  preceding  results.  Adding  these  four 
results,  the  sum  is  10,587,  which  is  the  sum  of  425,  36, 
9,215,  4,  and  907. 

NOTE.— It  frequently  happens,  when  adding  a  long  column  of  fig- 
ures, that  the  sum  of  two  numbers,  one  of  which  does  not  occur  in  the 
addition  table,  is  required.  Thus,  in  the  first  column  above,  the  sum  of 
16  and  6  was  required.  We  know  from  the  table  that  6  +  6  =  12; 
hence,  the  first  figure  of  the  sum  is  2.  Now,  the  sum  of  any  number  less 
than  20  and  of  any  number  less  than  10  must  be  less  than  thirty,  since 
20  +  10  =  30;  therefore,  the  sum  is  22.  Consequently,  in  cases  of  this 
kind,  add  the  first  figure  of  the  larger  number  to  the  smaller  number 
and,  if  the  result  is  greater  than  9,  increase  the  second  figure  of  the  larger 
number  by  1.  Thus,  44  +  7  =  ?  4  +  7  =  11;  hence,  44  +  7  =  51. 

3O.     The  addition  may  also  be  performed  as  follows: 

425 
36 

9215 

4 

907 


sum    10587    Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  units  column 
=  27  units,  or  2  tens  and  7  units.  Write  the  7  units  as  the 
first  or  right-hand  figure  in  the  sum.  Reserve  the  two  tens 
and  add  them  to  the  figures  in  tens  column.  The  sum  of 
the  figures  in  the  tens  column,  plus  the  2  tens  reserved  and 
carried  from  the  units  column  =  8,  which  is  written  down  as 
the  second  figure  in  the  sum.  There  is  nothing  to  carry  to 
the  next  column,  because  8  is  less  than  10.  The  sum  of  the 
numbers  in  the  next  column  is  15  hundreds,  or  1  thousand 
and  5  hundreds.  Write  down  the  5  as  the  third  or  hundreds 
figure  in  the  sum  and  carry  the  1  to  the  next  column.  1  -|- 
9  =  10,  which  is  written  down  at  the  left  of  the  other 
figures. 

The  second  method  saves  space  and  figures,  but  the  first 
js  to  be  preferred  when  adding  a  long  column. 


ARITHMETIC. 

31.     EXAMPLE. — Add  the  numbers  in  the  column  below. 
SOLUTION.—  890 

82 

90 
393 
281 

80 
770 

83 
492 

80 
383  " 

84 
191 


sum  3899  Ans. 

EXPLANATION. — The  sum  of  the  digits  in  the  first  column 
equals  19  units,  or  1  ten  and  9  units.  Write  down  the  9  and 
carry  1  to  the  next  column.  The  sum  of  the  digits  in  the 
second  column-)-  1  =  109  tens,  or  10  hundreds  and  9  tens. 
Write  down  the  9  and  carry  the  10  to  the  next  column.  The 
sum  of  the  digits  in  this  column  plus  the  10  reserved  =  38. 

The  entire  sum  is  3,899. 

32.  Rule  1. — (a)  Begin  at  the  right,  add  each  column 
separately,  and  write  the  sum,  if  it  be  only  one  figure,  under 
the  column  added. 

(6)  If  the  sum  of  any  column  consists  of  two  or  more  fig- 
ures, put  the  right-hand  figure  of  the  sum  under  that  column, 
and  add  t lie  remaining  figure  or  figures  to  the  next  column. 

33.  Proof. — To  prove  addition,  add  each  column  from 
top  to  bottom.     If  you  obtain  the  same  result  as  by  adding 
from  bottom  to  top,  the  work  is  probably  correct. 


EXAMPLES  FOR   PRACTICE. 

34.     Find  the  sum  of 

(a)  104  +  203  +  613  +  214.  \  (a)  1,134. 

(6)  1,875  +  3,143  +  5,826  +  10,832.       .    I  (d)  21,676. 

(0  4,865-+ 2,145 +  8,173 +  40,084.          "  1  (c)  55,267. 

(d)  14,204  +  8,173  +  1,065  +  10,043,  [  (d)  33,484. 


ARITHMETIC  9 

(*)    10,832  +  4,145  +  3,133  +  5,872.  f  (e)   23,982. 

(/)  214  +  1,231  +  141  +  5,000.  I    (/)  6,586. 

(g)  123  +  104  +  425  +  126  +  327.  Ans>   1(^)1,105. 

(K)   6,354  +  2,145  +  2,042  +  1,111  +  3,333.  [  (k)   14,985. 

1.  A  week's  record  of  coal  burned  in  an  engine  room  is  as  follows  : 
Monday,   1,800   pounds  ;    Tuesday,    1,655   pounds  ;  Wednesday,    1,725 
pounds;   Thursday,  1,690  pounds;   Friday,  1,648  pounds;   Saturday, 
1,020  pounds.     How  much  coal  was  burned  during  the  week  ? 

Ans.  9,538  pounds. 

2.  A  steam  pump  pumps  out  of  a  cistern  in  one  hour  4,200  gallons  ; 
in  the  next  hour,  5,420  gallons,  and  in  45  minutes  more,  an  additional 
3,600  gallons,  when  the  cistern  becomes  empty.     How  many  gallons 
were  in  the  cistern  at  first  ?  Ans.  13,220  gallons. 

3.  What  is  the  total  cost  of  a  steam  plant,  the  several  items  of 
expense  being  as  follows  :   Steam  engine,  $900  ;  boiler,  $775  ;  fittings 
and  connections,  $225  ;  erecting  the  plant,  $125  ;  engine  house,  $650  ? 

Ans.  $2,675. 


SUBTRACTION. 

35.  In  Arithmetic,  subtraction  is  the  process  of  find- 
ing how  much  greater  one  number  is  than  another. 

The  greater  of  the  two  numbers  is  called  the  minuend. 
The  smaller  of  the  two  numbers  is  called  the  subtrahend. 
The  number  left  after  subtracting  the  subtrahend  from  the 
minuend  is  called  the  difference    or  remainder. 

36.  The  sign  of   subtraction  is  — .     It  is  read  minus, 
and  means  less.     Thus,  12  —  7  is  read  12  minus  7,  and  means 
that  7  is  to  be  taken  from  12. 

37.  EXAMPLE.— From  7,568  take  3,425. 
SOLUTION.—  minuend  75  68 

subtrahend  3425 


remainder  4143    Ans. 

EXPLANATION. — Begin  at  the  right-hand  or  units  column 

and  subtract  in  succession  each  figure   in    the  subtrahend 

from  the  one  directly  above  it  in  the  minuend,  and  write 

the  remainders  below  the  line.     The   result  is  the   entire 

emainder. 


10  ARITHMETIC. 

38.  When  there  are  more  figures  in  the  minuend  than  in 
the  subtrahend,  and  when  some  figures  in  the  minuend  are 
less  than  the  figures  directly  under  them  in  the  sttbtrahend, 
proceed  as  in  the  following  example: 

EXAMPLE.— From  8,453  take  844. 

SOLUTION.—  minuend  8453 

subtrahend     844 


remainder  7609    Ans. 

EXPLANATION. — Begin  at  the  right-hand  or  units  column 
to  subtract.  We  can  not  take  4  from  3,  and  must,  therefore, 
borrow  1  from  5  in  tens  column  and  annex  it  to  the  3  in  units 
column.  The  1  ten  =  10  units,  which  added  to  the  3  in  the 
units  column  =  13  units.  4  from  13  =  9,  the  first  or  units 
figure  in  the  remainder. 

Since  we  borrowed  1  from  the  5,  only  4  remains ;  4  from  4 
—  0,  the  second  or  tens  figure.  We  can  not  take  8  from  4,  so 
borrow  1  thousand  or  10  hundreds  from  8;  10  hundreds  +  4 
hundreds  =  14  hundreds. 

8  from  14  =  6,  the  third  or  hundreds  figure  in  the  re- 
mainder. 

Since  we  borrowed  1  from  8  only  7  remains,  from  which 
there  is  nothing  to  subtract ;  therefore,  7  is  the  next  figure 
in  the  remainder  or  answer. 

The  operation  of  borrowing  is  placing  1  before  the  figure 
following  the  one  from  which  it  is  borrowed.  In  the  above 
example  the  one  borrowed  from  5  is  placed  before  3,  making 
it  13,  from  which  we  subtract  4.  The  1  borrowed  from  8  is 
placed  before  4,  making  14,  from  which  8  is  taken. 

39.      EXAMPLE.— Find  the  difference  between  10,000  and  8,763. 

SOLUTION. —  minuend  10000 

subtrahend       8763 


remainder     1237    Ans. 


EXPLANATION. — In  the  above  example  we  borrow  1  from 
the  second  column  and  place  it  before  0,  making  10;  3  from 
10  =  7\  In  the  same  way  we  borrow  1  and  place  it  before 


ARITHMETIC. 


11 


the  next  cipher,  making  10;  but  as  we  have  borrowed  1  from 
this  column  and  taken  it  to  the  units  column,  only  9  re- 
mains, from  which  to  subtract  0,  6 '  from  9  =  3.  For  the 
same  reason  we  subtract  7  from  9  and  8  from  9  for  the  next 
two  figures,  and  obtain  a  total  remainder  of  1,237.  • 

40.  Rule   2. — Place  the  subtralicnd  or  smaller  number 
under  the  minuend  or  larger  number,  in  the  same  manner  as 
for  addition,  arid  proceed  as  in  Arts.  37,  38,  and  39. 

41.  Proof. —  To  prove  an  example  in   subtraction,  add 
the  subtrahend  and  remainder.      The  sum  sJwuld  equal  the 
minuend.     If  it  does  not,  a  mistake  has  been  made,  and  the 
work  should  be  done  over. 

Proof  of  the  above  example: 

subtrahend     8703 
remainder      1237 


minuend  10000 


EXAMPLES  FOR  PRACTICE. 
42.      From 

(a)  94,278  take  62,574. 

(b)  53,714  take  25,824. 

(c)  71,832  take  58,109. 

(d)  20,804  take  10,408.  An 

(e)  310,465  take  102,141. 
(/)(81,043  +  1,041)  take  14,831. 
(^) (20,482 +  18,216)  take  21,214. 
(//)  (2,040  +1,213  +  542)  take  3,791. 


(a)  31,704. 

(b)  27,890. 
(f)  13,723. 

(d)  10,396. 

(e)  208,324. 
(/)  67,253. 
(£•)  17,484. 


1.  A  cistern  is  fed  by  two  pipes  which  supply  1,200  and  2,250  gallons 
per  hour,  respectively,  and  is  being  emptied  by  a  pump  which  delivers 
5.800  gallons  per  hour.     Starting  with  8,000  gallons  in  the  cistern,  how 
much  water  does  it  contain  at  the  end  of  an  hour  ?   Ans.  5,650  gallons. 

2.  A  train  in  running  from  New  Vork  to  Buffalo  travels  38  miles 
the  first  hour,  42  the  second,  39  the  third,  56  the  fourth.  52  the  fifth, 
and  48  the  sixth  hour.     How  many  miles  remain  to  be  traveled  at  the 
end  of  the  sixth  hour,  the  distance  between  the  two  places  being  410 
miles?  Ans.  135  miles. 


12  ARITHMETIC. 

3.  On  Monday  morning  a  bank  had  on  hand  $2,862.  During  the 
day  $1,831  were  deposited  and  §2,172  drawn  out;  on  Tuesday,  $3,126 
were  deposited,  and  §1,954  drawn  out.  How  many  dollars  were  on 
hand  Wednesday  morning  ?  Ans.  §3,693. 


MULTIPLICATION. 

43.  To  multiply  a  number  is  to  add  it  to  itself  a  cer- 
tain number  of  times. 

44.  Multiplication  is  the  process  of  multiplying  one 
number  by  another. 

The  number  thus  added  to  itself,  or  the  number  to  be 
multiplied,  is  called  the  multiplicand. 

The  number  which  shows  how  many  times  the  multiplicand 
is  to  be  taken,  or  the  number  by  which  we  multiply,  is  called 
the  multiplier. 

The  result  obtained  by  multiplying  is  called  the  product. 

45.  The  sign  of  multiplication  is  X .     It  is  read  times 
or  multiplied  by.     Thus,  9  X  6  is  read  9  times  6,  or  9  multi- 
plied by  6. 

46.  It  matters  not  in  what  order  the  numbers  to  be 
multiplied  together  are  placed.     Thus,  6  X  9  is  the  same  as 
9X6. 

47.  In  the  following  table,  the  product  of  any  two  num- 
bers (neither  of  which  exceeds  twelve)  may  be  found: 


ARITHMETIC. 


13 


TABLE    2. 


1  times    1  is        1 
1  times    2  are      2 
1  times    3  are      3 
1  times    4  are      4 
1  times    5  are      5 
1  times    6  are      6 
1  times    7  are      7 
1  times    8  are      8 
1  times    9  are      9 
1  times  10  are    10 
1  times  11  are    11 
1  times  12  are    12 

2  times    1  are      2 
2  times    2  are      4 
2  times    3  are      6 
2  times    4  are      8 
2  times    5  are    10 
2  times    6  are    12 
2  times    7  are    14 
2  times    8  are    16 
2  times    9  are    18 
2  times  10  are    20 
2  times  11  are    22 
2  times  12  are    24 

3  times    1  are      3 
3  times    2  are      6 
3  times    3  are      9 
3  times    4  are    12 
3  times    5  are    15 
3  times    6  are    18 
3  times    7  are    21 
3  times    8  are    24 
3  times    9  are    27 
3  times  10  are    30 
3  times  11  are    33 
3  times  12  are    36 

4  times    1  are      4 
4  times    2  are      8 
4  times    3  are    12 
4  times    4  are    16 
4  times    5  are    20 
4  times    6  are    24 
4  times    7  are    28 
4  times    8  are    32 
4  times    9  are    36 
4  times  10  are    40 
4  times  11  are    44 
4  times  12  are    48 

5  times    1  are      5 
5  times    2  are    10 
5  times    3  are    15 
5  times    4  are    20 
5  times    5  are    25 
5  times    6  are    30 
5  times    7  are    35 
5  times    8  are    40 
5  times    9  are    45 
5  times  10  are    50 
5  times  11  are    55 
5  times  12  are    60 

6  times    1  are      6 
6  times    2  are    12 
6  times    3  are    18 
6  times    4  are    24 
6  times    5  are    30 
6  times    6  are    36 
6  times    7  are    42 
6  times    8  are    48 
6  times    9  are    54 
6  times  10  are    60 
6  times  11  are    66 
6  times  12  are    72 

7  times    1  are     7 
7  times    2  are    14 
7  times    3  are    21 
7  times    4  are    28 
7  times    5  are    35 
7  times    6  are   42 
7  times    7  are    49 
7  times    8  are    56 
7  times    9  are    63 
7  times  10  are    70 
7  times  11  are    77 
7  times  12  are    84 

8  times    1  are      8 
8  times    2  are    16 
8  times    3  are    24 
8  times    4  are    32 
8  times    5  are    40 
8  times    6  are    48 
8  times    7  are    56 
8  times    8  are    64 
8  times    9  are    72 
8  times  10  are    80 
8  times  11  are    88 
8  times  12  are    96 

9  times    1  are      9 
9  times    2  are    18 
9  times    3  are    27 
9  times    4  are    36 
9  times    5  are    45 
9  times    6  are    54 
9  times    7  are    63 
9  times    8  are    72 
9  times    9  are    81 
9  times  10  are    90 
9  times  11  are    99 
9  times  12  are  108 

10  times    1  are    10 
10  times    2  are    20 
10  times    3  are    30 
10  times    4  are    40 
10  times    5  are    50 
10  times    6  are    60 
10  times    7  are    70 
10  times    8  are    80 
10  times    9  are    90 
10  times  10  are  100 
10  times  11  are  110 
10  times  12  are  120 

11  times    1  are    11 
11  times    2  are    22 
11  times    3  are    33 
11  times    4  are    44 
11  times    5  are    55 
11  times    6  are    66 
11  times    7  are    77 
11  times    8  are    88 
11  times    9  are    99 
11  times  10  are  110 
11  times  11  are  121 
11  times  12  are  132 

12  times    1  are    12 
12  times    2  are    24 
12  times    3  are    36 
12  times    4  are    48 
12  times    5  are    60 
12  times    6  are    72 
12  times    7  are    84 
12  times    8  are    96 
12  times    9  are  108 
12  times  10  are  120 
12  times  11  are  132 
12  times  12  are  144 

This  table  should  be  carefully  committed  to  memory. 
Since  0  has  no  value,  the  product  of  0  and  any  number  is  0. 


14  ARITHMETIC. 

48.     To  multiply  a  number  by  one  figure  only : 

EXAMPLE.— Multiply  425  by  5. 
SOLUTION. —  multiplicand       425 

multiplier  5 

product    2125    Ans. 

EXPLANATION. — For  Convenience,  the  multiplier  is  gener- 
ally written  under  the  rigkt-Jiand  figure  of  the  multiplicand. 
On  looking  in  the  multiplication  table,  we  see  that  5x5  are  25. 
Multiplying  the  first  figure  at  the  right  of  the  multiplicand, 
or  5,  by  the  multiplier  5,  it  is  seen  that  5  times  5  units  are  25 
units,  or  2  tens  and  5  units.  Write  the  5  units  in  units  place 
in  the  product,  and  reserve  the  2  tens  to  add  to  the  product  of 
tens.  Looking  in  the  multiplication  table  again,  we  see  that 
5x2  are  10.  Multiplying  the  second  figure  of  the  multipli- 
cand by  the  multiplier  5,  we  see  that  5  times  2  tens  are  10 
tens,  plus  the  2  tens  reserved,  are  12  tens,  or  1  hundred  plus 
2  tens.  Write  the  2  tens  in  tens  place,  and  reserve  the  1  hun- 
dred to  add  to  the  product  of  hundreds.  Again,  we  see  by 
the  multiplication  table  that  5x4  are  20.  Multiplying  the 
third  or  last  figure  of  the  multiplicand  by  the  multiplier  5,  we 
see  that  5  times  4  hundreds  are  20  hundreds,//^  the  1  hun- 
dred reserved,  are  21  hundreds,  or  2  thousands //«.$•  1  hundred, 
which  we  write  in  thousands  and  hundreds  places,  respectively. 
Hence,  \)i\&  product  is  2,125. 
This  result  is  the  same  as  adding  425  five  times.  Thus, 

425 

425 

425 

425 

425 

sum    2125    Ans. 


EXAMPLES  FOR  PRACTICE. 

49.      Find  the  product  of 

(a)     61,483X6.  (  (a)  368,898. 

(6)     12,375x5.  A  =   J    W  61,875. 

(f)     10,426  X  7.  ~'   \    (f)  72,982. 

(<0    10,835x3.  [  (d)  32,505. 


ARITHMETIC.  16 

(e)     98,376x4  f   (e)  393,504. 

(/)    10,873X8.  (/)  86,984. 

(g)    71,543X9.  1  (^)  643,887. 

(A)    218,734X2.  [   (A)  437,468. 

1.  A  stationary  engine  makes  5,520  revolutions  per  hour.     Running 
9  hours  a  day,  5  days  in  the  week,  and  5  hours  on  Saturday,  how  many 
revolutions  would  it  make  in  4  weeks  ?  Ans.  1,104,000  revolutions. 

2.  An  engineer  earns  $650  a  year,  and  his  average  expenses  are 
$548.     How  much  could  he  save  in  8  years  at  that  rate  ?         Ans.  $816. 

3.  The  connection  between  an  engine  and  boiler  is  made  up  of  5 
lengths  of  pipe,  three  of  which  are  12  feet  long,  one  2  feet  6  inches 
long,  and  one  8  feet  6  inches  long.     If  the  pipe  weighs  9  pounds  per 
foot,  what  is  the  total  weight  of  the  pipe  used  ?  Ans.  423  pounds. 

5O.     To  multiply  a  number  by  two  or  more  fig- 
ures : 

EXAMPLE. — Multiply  475  by  234. 
SOLUTION.—        multiplicand  475 

multiplier  234 

1900 
1425 
950 


product    111150    Ans. 

EXPLANATION. — For  convenience,  the  multiplier  is  gener- 
ally written  under  the  multiplicand,  placing  units  under 
units,  tens  under  tens,  etc. 

We  can  not  multiply  by  234  at  one  operation ;  we  must, 
therefore,  multiply  by  the  parts  and  then  add  the  partial 
products. 

The  parts  by  which  we  are  to  multiply  are  4  units,  3  tens, 
and  2  hundreds.  4  times  475  =  1,900,  the  first  partial  prod- 
uct;  3  times  475  =  1,425,  the  second  partial  product,  the 
right-hand  figure  of  which  is  written  directly  under  the  fig- 
ure multiplied  by,  or  3 ;  2  times  475  =  950,  the  third  partial 
product,  the  right-hand  figure  of  which  is  written  directly 
under  the  figure  multiplied  by,  or  2. 

The  sum  of  these  three  partial  products  is  111,150,  which 
is  the  entire  product. 


16  ARITHMETIC. 

5 1 .  Rule  3. — (a)  Write  the  multiplier  under  the  multipli- 
cand, so  that  units  are  -under  units,  tens  under  tens,  etc. 

(6)  Begin  at  the  right  and  multiply  each  figure  of  the  multi- 
plicand by  each  successive  figure  of  the  multiplier,  placing  the 
right-hand  figure  of  each  partial  product  directly  under  the 
figure  used  as  a  multiplier, 

(c)  The  sum  of  the  partial  products  will  equal  the  required 
product. 

52.  Proof. — Review  the  work  carefully,  or  multiply  the 
multiplier  by  the  multiplicand ;  if  the  results  agree,  the  work 
is  correct. 

53.  When  there  is  a  cipher  in  the  multiplier,  multiply 
the  entire  multiplicand  by  it ;  since  the  result  will  be  zero, 
place  a  cipher  under  the  cipher  in  the  multiplier.     Thus, 

(a)  (6)  (c)  (d) 

0  2  15  708 

XO  XO  X     0  X       0 

0    Ans.        0    Ans.       ~0~  Ans.  0    Ans. 

to  CO  (JO 

3114  4008  31264 

203  305  1002 


9342  20040  62528 

120240          3126400 


632142  Ans,    1222440  Ans.    31326528  Ans. 

In  examples  (e],  (/),  and  (g),  we  multiply  by  0  as  directed 
above;  then  multiply  by  the  next  figure  of  the  multiplier 
and  place  the  first  figure  of  the  product  alongside  the  0,  as 
shown. 


EXAMPLES  FOR  PRACTICE. 

54.     Find  the  product  of 
(a)     3,842  X  26. 


(b)  3,716  X  45. 

(c)  1,817  X  124 

(d)  675X38. 


Ans. 


(b)  167,220. 

(c)  225,308. 

(d)  25,650. 


ARITHMETIC. 

(*)      1,875x33.  r    (e)    61,875. 

(/)    4,836X47.  I    (/)    227,292. 


(.£-)  5,682x543. 

(>&)  3,257  X  246. 

(/)  2,875  X  302. 

(/)  17,819  X  1,004.  Ans. 

(k)  38,674  X  205. 

(/)  18,304x100. 

(m)  7,834  X  10. 

(«)  87,543x1,000. 

(o)  48,763  X  100. 


(£•)  3,085,326. 

(h)  801,222. 

(/)  868,250. 

(f)  17,890,276. 

(k)  7,928,170. 

(/)  1,830,400. 

(m)  78,340. 

(«)  87,543,000. 

(o)  4,876,300. 


1.  If  the  area  of  a  steam-engine  piston  is  113  square  inches,  what  is 
the  total  pressure  upon  it  when  the  steam  pressure  is  85  pounds  per 
square  inch  ?  Ans.  9,605  pounds. 

2.  A  steam  engine,  which  indicated  164  horsepower,  was  found  to 
consume  4  pounds  of  coal  per  horsepower  per  hour.     Being  replaced 
by  a  new  engine,  which  was  of  the  same  horsepower  as  the  other, 
another  test  was  made,  which  showed  a  consumption  of  3  pounds  per 
horsepower  per  hour.     What  was  the  saving  of  coal  for  a  year  of  309 
days,  if  the  engine  averaged  to  run  14  hours  a  day  ? 

Ans.  709,464  pounds. 

3.  Two  steamers  are  7,846  miles  apart,  and  are  sailing  towards  each 
other,  one  at  the  rate  of  18  miles  an  hour,  and  the  other  at  the  rate  of 
15  miles  an  hour.     How  far  apart  will  they  be  at  the  end  of  205  hours  ? 

Ans.  1,081  miles. 

DIVISION. 

55.  Division  is  the  process  of  finding  how  many  times 
one  number  is  contained  in  another  of  the  same  kind. 

The  number  to  be  divided  is  called  the  dividend. 
The  number  by  which  we  divide  is  called  the  divisor. 
The  number  which  shows  how  many  times  the  divisor  is 
contained  in  the  dividend  is  called  the  quotient. 

56.  The  sign  of  division  is  -=-.     It  is  read  divided  by. 
54  -T-  9  is  read  54  divided  by  9.     Another  way  to  write  54 

divided  by  9  is  ^.      Thus,  54  ^  9  =  6,  or  ~  =  6. 
y  y 

In  both  of  these  cases  54  is  the  dividend  and  9  is  the 
divisor. 

Division  is  the  reverse  of  multiplication. 


18  ARITHMETIC. 

57.  To  divide  when  the  divisor  consists  of  but 
one  figure,  proceed  as  in  the  following  example : 

EXAMPLE.— What  is  the  quotient  of  875  -H  7  ? 
divisor  dividend  quotient 

SOLUTION. —  7)875(125    Ans. 

7_ 
17 
14 
35 
35 
remainder     0 

EXPLANATION.^ — 7  is  contained  in  8  hundreds  1  hundred 
times.  Place  the  one  as  the  first,  or  left-hand,  figure  of  the 
quotient.  Multiply  the  divisor  7  by  the  1  hundred  of  the 
quotient,  and  place  the  product  7  hundreds  under  the 
8  hundreds  in  the  dividend,  and  subtract.  Beside  the 
remainder  1,  bring  down  the  7  tens,  making  17  tens;  17 
divided  by  7  =  2  times.  Write  the  two  as  the  second 
figure  of  the  quotient.  Multiply  the  divisor  7  by  the  2, 
and  subtract  the  product  from  17.  Beside  the  remain- 
der 3,  bring  down  the  5  units  of  the  dividend,  making 
35  units.  7  is  contained  in  35,  5  times,  which  is  placed  in 
the  quotient.  Multiplying  the  divisor  by  the  last  figure  of 
the  quotient,  5  times  7  =  35,  which  subtracted  from  35, 
under  which  it  is  placed,  leaves  0.  Therefore,  the  quotient 
is  125.  This  method  is  called  long  division. 

58.  In  short  division,  only  the  divisor,  dividend,  and 
quotient  are  written,  the  operations  being  performed  men- 
tally. 

dividend 

divisor    7  )  8  ]  7 3  5 

quotient    125     Ans. 

The  mental  operation  is  as  follows:  7  is  contained  in  8, 
once  and  one  remainder;  1  placed  before  7  makes  17;  7  is 
contained  in  17,  2  times  and  3  over;  the  3  placed  before  5 
makes  35 ;  7  is  contained  in  35,  5  times.  These  partial  quo- 
tients placed  in  order  as  they  are  found,  make  the  entire 
quotient  125. 


ARITHMETIC.  19 

The  small  figures  are  placed  in  the  example  given  to  better 
illustrate  the  explanation;  they  are  never  written  when 
actually  performing  division  in  this  way. 

59.     If  the  divisor  consists  of  2  or  more  figures,  proceed 
as  in  the  following  example : 
EXAMPLE.— Divide  2,702,826  by  63. 

divisor  dividend          quotient 

SOLUTION.—  63  )2702826(42902    Ans. 

252 


182 
126 

568 
567 


126 
126 


EXPLANATION. — As  63  is  not  contained  in  the  first  two  fig- 
ures, 27,  we  must  use  the  first  three  figures,  270.  Now,  by 
trial,  we  must  find  how  many  times  63  is  contained  in  270; 
6  is  contained  in  the  first  two  figures  of  270,  4  times.  Place 
the  4  as  the  first  or  left-hand  figure  in  the  quotient.  Multi- 
ply the  divisor  63  by  4,  and  subtract  the  product  252  from 
270.  The  remainder  is  18,  beside  which  we  write  the  next 
figure  of  the  dividend,  2S  making  182.  Now,  6  is  contained 
in  the  first  two  figures  of  182,  3  times,  but  on  multiplying 
63  by  3,  we  see  that  the  product  189  is  too  great,  so  we  try 
2  as  the  second  figure  of  the  quotient.  Multiplying  the 
divisor  63  by  2,  and  subtracting  the  product  126  from  182, 
the  remainder  is  56,  beside  which  we  bring  down  the  next 
figure  of  the  dividend,  making  568;  6  is  contained  in  56 
about  9  times.  Multiply  the  divisor  63  by  9  and  subtract 
the  product  567  from  568.  The  remainder  is  1,  and  bringing 
down  the  next  figure  of  the  dividend,  2,  gives  12.  As  12  is 
smaller  than  63,  we  write  0  in  the  quotient  and  bring  down 
the  next  figure,  6,  making  126.  63  is  contained  in  126,  2 
times,  without  a  remainder.  Therefore.  42,902  is  the  quo- 
tient. 


20  ARITHMETIC. 

6O.  Rule  4.—  (a)  Write  the  divisor  at  the  left  of  the 
dividend,  with  a  line  between  them. 

(bi\     Find  Jiow  many  times  the  divisor  is  contained  in  the 

lowest    number  of   the    left-hand  figures  of    the    dividend 

•  that  will  contain  it,  and  write  the  result  at  the  right  of  the 

dividend,   with  a   line   between,  for   the  first  figure  of  the 

quotient. 

(c)  Multiply  the  divisor  by  this  quotient  ;  write  the  product 
tinder  the  partial  dividend  used,  and  subtract,  annexing  to  the 
remainder  the  next  figure  of  the  dividend.     Divide  as  before, 
and  thus  continue  until  all  the  figures  of  the  dividend  have 
been  used. 

(d)  If  any  partial  dividend  will  not  contain  the  divisor, 
write  a  cipher  in  the  quotient,  annex  the  next  figure  of  the 
dividend  and  proceed  as  before. 

(e)  If  there  be  a  remainder  at  last,  write  it  after  the  quo- 
tient, with  the  divisor  underneath. 

61  .  Proof.  —  Multiply  the  quotient  by  the  divisor,  and  add 
the  remainder,  if  there  be  any,  to  the  product.  The  result  will 
be  the  dividend. 

divisor  dividend  quotient 

Thus,  63)4235(67£|    Ans. 

378 

~455 
441 

remainder  1  4 

Proof,  quotient  6  7 

divisor  6  3 


402 

*    4221 
remainder  1  4 

dividend      4235 


ARITHMETIC.  21 

EXAMPLES  FOR   PRACTICE. 

62*      Divide  the  following: 

(a)   126,498  by  58.  (a)   2,181. 

(d)    3,207,594  by  767.  (i,)   4,182. 

(c)  11,408,202  by  234.  (c)   48,753. 

(d)  2,100,315  by  581.  (d)  3,615. 
(<?)    969,936  by  4,008.                                Ans-  \     (e)    242. 
(/)   7,481,888  by  1,021.                                           (/)  7,328. 
(g)  1,525,915  by  5,003.                                         (g)  305. 

(A)    1,646,301  by  381.  [   (k)   4,321. 

1.  In  a  mile  there  are  5,280  feet.     How  many  rails  would  it  take  to 
lay  a  double  row  one  mile  long,  each  rail  being  30  feet  long  1 

Ans.  352  rails. 

2.  How  many  rivets  will  be  required  for  the  longitudinal  seams  of 
a  cylindrical  boiler  20  feet  long,  the  joint  being  double  riveted,  and  the 
rivets  being  spaced  4  inches  apart  1  Ans.  120  rivets. 

NOTE. — First  find  the  length  of  the  boiler  in  inches. 

3.  It  requires  7,020,000  bricks  to  build  a  large  foundry.     How  many 
teams  will  it  require  to  draw  the  bricks  in  60  days,  if  each  team  draws 
6  loads  per  day  and  1,500  bricks  at  a  load  ?  Ans.  13  teams. 

NOTE.— Find   how  many  loads  7,020,000  bricks  make  ;    then,  how 
many  days  it  will  take  one  team  to  draw  the  brick. 


CANCELATION. 

63.  Cancelation    is  the  process  of  shortening  opera- 
tions in  division   by  casting  out  equal   factors'  from  both 
dividend  and  divisor. 

64.  The  factors  of  a  number  are  those  numbers  which, 
when  multiplied  together,  will  equal  that  number.     Thus,  5 
and  3  are  factors  of  15,  since  5x3  =  15.     Likewise,  8  and  7 
are  the  factors  of  50,  since  8  X  7  =  56. 

65.  A  prime  number  is  one  which  can  not  be  divided 
by  any  number  except  itself  and  1.     Thus,  2,  3,  11,  29,  etc., 
are  prime  numbers. 

66.  A    prime    factor    is  any  factor  that  is  a   prime 
number. 

Any  number  that  is  not  a  prime  is  called  a  composite 
number,  and  may  be  produced  by  multiplying  together  its 
prime  factors.  Thus,  GO  is  a  composite  number,  anc}  is  equal 
to  the  product:  of  its  prime  factors,  2  X  3  X  3  X  5, 


22  ARITHMETIC. 

Numbers  are  said  to  be  prime  to  each  other  when  no 
two  of  them  can  be  divided  by  any  number  except  1 ;  the 
numbers  themselves  may  be  either  prime  or  composite. 
Thus,  the  numbers  3,  5,  and  11  are  prime  to  each  other,  so 
also  are  22,  25,  and  21 — all  composite  numbers. 

67.  Canceling  equal  factors  from  both  dividend  and  divi- 
sor does  not  change  the  quotient. 

The  canceling  of  a  factor  in  botJi  dividend  and  divisor  is  the 
same  as  dividing  them  both  by  the  same  number,  which,  by 
the  principle  of  division,  does  not  change  the  quotient. 

Write  the  numbers  which  make  the  dividend  above  the  line,, 
and  those  which  make  the  divisor  below  it. 

68.  EXAMPLE.— Divide  4  X  45  X  60  by  9  X  24. 

SOLUTION. — Placing  the  dividend  over  the  divisor,  and  canceling 

Ans. 


1 

EXPLANATION. — The  4  in  the  dividend  and  24  in  the  divisor 
are  both  divisible  by  4,  since  4  divided  by  4  equals  1,  and  24 
divided  by  4  equals  6.  Cross  off  the  four  and  write  the  1 
over  it;  also,  cross  off  the  24  and  write  the  6  under  it.  Thus, 


60  in  the  dividend  and  G  in  the  divisor  are  divisible  by  6, 
since  60  divided  by  6  equals  10,  and  6  divided  by  6  equals  1. 
Cross  off  the  60  and  write  10  over  it  ;  also,  cross  off  the  6  and 
write  1  under  it.  Thus, 


Again,  45  in  the  dividend  and  9  in  the  divisor  are  divisible 
by  9,  since  45  divided  by  9  equals  5,  and  9  divided  by  9  equals 
1.  Cross  off  the  45  and  write  the  5  over  it  ;  also,  cross  off  the 
9  and  write  the  1  under  it.  Thus, 

l      5'     10 

I  x 


ARITHMETIC.  23 

Since  there  are  no  two  remaining  numbers  (one  in  the  divi- 
dend and  one  in  the  divisor)  divisible  by  any  number  except 
1,  without  a  remainder,  it  is  impossible  to  cancel  further. 

Multiply  all  the  uncanceled  numbers  in  the  dividend 
together,  and  divide  their  product  by  the  product  of  all  the 
uncanceled  numbers  in  the  divisor.  The  result  will  be  the 
quotient.  The  product  of  all  the  uncanceled  numbers  in 
the  dividend  equals  5  X  1  X  10  =  50;  the  product  of  all  the 
uncanceled  numbers  in  the  divisor  equals  1x1  =  1. 

1      5      10 

£  X  &  X  W    1  x  5  x  10      _A 

Hence-     yx#     =     ixi      =5°-    Ans' 
l       0 

It  is  usual  to  omit  the  1's  when  canceling  them,  instead 
of  writing  them  as  above. 

69.  Rule  5. — (a)  Cancel  the  common  factors  from  both 
the  dividend  and  divisor, 

(b)  Then  divide  the  product  of  the  remaining  factors  of  the 
dividend  by  the  product  of  the  remaining  factors  of  the  divisor, 
and  the*result  will  be  the  quotient. 


EXAMPLES  FOR   PRACTICE. 
7O.      Divide 

(a)   14  X  18  X  16  X  40  by  7  X  8  X  6  X  5  X  3. 
(&)    3  X  65  X  50  X  100  X  60  by  30  X  60  X  13  X  10. 
(<•)    8  X  4  X  3  X  9  X  11  by  11  X  9  X  4  X  3  X  8- 
00   164  X  321  X  6  X  7  X  4  by  82  X  321  X  7. 
(e)    50  X  100  X  200  X  72  by  1,000  X  144  X  100. 
(/)  48  X  63  X  55  X  49  by  7  X  21  X  11  X  48. 
(g )  110  X  150  X  84  X  32  by  11  X  15  X  100  X  64. 
(A)    115  X  120  X  400  X  1,000  by  23  X  1,000  X  60  X  800. 


(a)  32. 
(6)  250. 

(c)  1. 
OO  48. 

(e)  5. 
(/)  105. 
(£0  42. 

(k)  5. 


FRACTIONS. 

71.  A  fraction  is  a  part  of  a  whole  number:  One-half, 
one-third,  two-fifths  are  fractions. 

72.  Two  numbers  are  required  to  express  a  fraction, 
one   called  the  numerator,    and  the   other  the  denomi- 
nator. 


24  ARITHMETIC. 

73.  The  numerator  is   placed   above   the   denominator, 
with  a  line  between  them ;  as,  f .     3  is  the  denominator,  and 
shows  into  how  many  equal  parts  the  unit  or  0«^  is  divided. 
The  numerator  2  shows  how  many  of  these  equal  parts  are 
taken  or   considered.     The  denominator  also  indicates  the 
names  of  the  parts. 

£  is  read  one-half. 

•f-  is  read  three-fourths. 

|  is  read  three-eighths. 

-£$  is  read  five-sixteenths. 

|4  is  read  twenty-nine-forty-sevenths. 

In  the  expression  "T  of  an  apple,"  the  denominator 
4  shows  that  the  apple  is  to  be  (or  has  been)  cut  into  4  equal 
parts,  and  the  numerator  3  shows  that  three  of  these  parts, 
or  four -ti ks,  are  taken  or  considered. 

If  each  of  the  parts,  or  fourths,  of  the  apple  were  cut  in 
two  equal  pieces,  there  would  then  be  twice  as  many  pieces 
as  before,  or  4x2  =  8  pieces  in  all;  one  of  these  pieces  would 
be  called  one-eighth,  and  would  be  expressed  in  figures  as  \. 
Three  of  these  pieces  would  be  called  three-eighths,  and 
written  -f.  The  words  three-fourths,  three-eighths,  five- 
sixteenths,  etc.,  are  abbreviations  of  three  one-fourths,  three 
one-eighths,  five  one-sixteenths,  etc.  It  is  evident  that  the 
larger  the  denominator,  the  greater  is  the  number  of  parts 
into  which  anything  is  divided;  consequently,  the  parts 
themselves  are  smaller,  and  the  value  of  the  fraction  is  less 
for  the  same  number  of  parts  taken.  In  other  words,  £,  for 
example,  is  smaller  than  |,  because  if  an  object  be  divided 
into  9  parts,  the  parts  are  smaller  than  if  the  same  object 
had  been  divided  into  8  parts ;  and,  since  £  is  smaller  than  |, 
it  is  clear  that  7  one-ninths  is  a  smaller  amount  than  7  one- 
eighths.  Hence,  also,  £  is  less  than  f . 

74.  The  value  of  a  fraction  is  the  result  obtained  by 
dividing  the  numerator  by  the  denominator;  as,  £  =  2,  |  =  3. 

75.  The  line  between  the  numerator  and  denominator 
means  divided  by,  or  -=-. 

f  is  equivalent  to  3  -4-  4. 
|  is  equivalent  to  6  -r  8, 


ARITHMETIC.  25 

76.  The  numerator  and  denominator  of  a  fraction  are 
called  the  terms  of^a  fraction. 

77.  The  value  of  a  fraction  whose  numerator  and  de- 
nominator are  equal,  is  1. 

f,  or  four-fourths,  =1. 
f,  or  eight-eighths,  =  1. 
-§-£•,  or  sixty-four  sixty-fourths,  =  1. 

78.  A  proper  fraction  is  a  fraction  whose  numerator 
is  less  than  its  denominator.      Its  value  is  /?.«  than  1 ;  as,  £ , 

f ,  iV 

79.  An  improper  fraction  is  a  fraction  whose  numer- 
ator equals  or  is  greater  than  the  denominator.     Its  value  is 
0«^  or  *«07V  than  0«^;  as,  £,  -|,  £f . 

80.  A  mixed  number  is  a  whole  number  and  a.  frac- 
tion united.     4f  is  a  mixed  number,   and  is  equivalent  to 
4  +  f.     It  is  read  four  and  two-thirds. 


REDUCTION    OF    FRACTIONS. 

81.  Reduction  of  fractions  is  the  process  of  chang- 
ing their  form  without  changing  their  value. 

82.  A  fraction  is  reduced  to  higher  terms  by  multiplying 
both  terms  of  the  fraction  by  the  same  number.     Thus,  f  is 
reduced  to  -f  by  multiplying  both  terms  by  2. 

3_X2  =6 
4X2       8* 

The  value  is  not  changed,  since  f  =  f .  For,  suppose  that 
an  object,  say  an  apple,  is  divided  into  8  equal  parts.  If 
these  parts  be  arranged  into  4  piles,  each  containing  2  parts, 
it  is  evident  each  pile  will  be  composed  of  the  same  amount 
of  the  entire  apple  as  would  have  been  the  case  had  the 
apple  been  originally  cut  into  4  equal  parts.  Now,  if  one  of 
these  piles  (containing  2  parts)  be  removed,  there  will  be  3 
piles  left  each  containing  2  equal  parts,  or  C  equal  parts  in 
all,  i.  e.,  six-eighths.  But,  since  one  pile,  or  one  quarter, 
was  removed,  there  are  three-quarters  left.  Hence,  f  =  \. 


26  ARITHMETIC. 

The  same  course  of  reasoning  may  be  applied  to  any  similar 
case.  Therefore,  multiplying  both  terms  of  a  fraction  by 
the  same  number  does  not  alter  its  value. 

3  X  2_  6 

4  X  2  ~  8' 

83.  A  fraction  is  reduced  to  lower   terms  by  dividing 
both  terms  by  the  same  number.    Thus,  -^  is  reduced  to  \  by 
dividing  both  terms  by  2. 

_8_-*-  2_  4 

10  -7T    3  1£ 

That  -^  =  -f  is  readily  seen  from  the  explanation  given  in 
Art.  82 ;  for,  multiplying  both  terms  of  the  fraction  -f  by  2, 
ix  1  =  T8<r»  and,  if  I  =  T8<r>  TSTT  must  equal  i  Hence,  dividing 
both  terms  of  a  fraction  by  the  same  number  does  not  alter 
its  value. 

84.  A  fraction  is  reduced  to  its  lowest  terms  when  its 
numerator  and  denominator  can  not  be  divided  by  the  same 
number,  as  f,  f ,  -f^-. 

85.  To  reduce  a  whole  number  or  mixed  num- 
ber to  an  improper  fraction  : 

EXAMPLE. — How  many  fourths  in  5  ? 

SOLUTION.— Since  there  are  4 fourths  in  1  (f  =  IV  in  5  there  will  be 
5x4  fourths,  or  20  fourths,  5  X  |  =  -^  Ans. 

EXAMPLE. — Reduce  8f  to  an  improper  fraction. 
SOLUTION.  —  8  x  f  =  ^.     \8-  +  f  =  ^.     Ans. 

86.  Rule  6. — Multiply  the  whole  number  by  the  denom- 
inator of  the  fraction,  add  the  numerator  to  the  product,  and 
place  the  denominator  under  the  result. 


EXAMPLES  FOR   PRACTICE. 

87.     Reduce  to  improper  fractions: 

(a) 

W  10TV 
(<0  87*. 
(*5  50f 


ARITHMETIC.  27 

88.  To  reduce  an  improper  fraction  to  a  whole 
or  mixed  number : 

EXAMPLE. — Reduce  ^  to  a  mixed  number. 

SOLUTION. — 4  is  contained  in  21,  5  times  and  1  remaining;  as  this  is 
also  divided  by  4,  its  value  is  £.  Therefore,  5  +  ±,  or  5£,  is  the  num- 
ber. Ans. 

89.  Rule  7. — Divide  the  numerator  by  the  denominator; 
the  quotient  will  be  the  wJiolc   number;   the   remainder,    if 
there  be  any,  will  be  the  numerator  of  the  fractional  part  of 
which  the  denominator  is  the  same  as  the  denominator  of  the 
improper  fraction. 

EXAMPLES  FOR  PRACTICE. 

90.  Reduce  to  whole  or  mixed  numbers  : 
(a)  HA- 

'       SI 

(')   *f 

(/)  W-  

91.  A  common  denominator  of  two  or  more  frac- 
tions is  a  number  which  will  contain  all  of  the  denominators 
of  the  fraction s  without  a  remainder.     The  least  common 
denominator  is  the  least  number  that  will  contain  all  of 
the  denominators  of  the  fractions  without  a  remainder. 

92.  To  find  the  least  common  denominator: 

EXAMPLE. — Find  the  least  common  denominator  of  ^,  \,  \,  and  -j>ff. 
SOLUTION. — We  first  place  the  denominators  in  a  row,  separated  by 
commas. 

2  )  4,  3,  9,  16 

2  )  2,  3,  9.    8 

3  )  1,  3,  9,    4 
8  )  1,  1,  8,    4 
4)1.  1,  1,    4 

1,  1,  1,    1 

2x2x3x3x4  =  144,  the  least  common  denominator.     Ans. 

EXPLANATION. — Divide  each  of  them  by  some  prime  num- 
ber which  will  divide  at  least  two  of  them  without  a  remain- 
der (if  possible),  bringing  down  those  denominators  to  the  row 


28  ARITHMETIC. 

below  which  will  not  contain  the  divisor  without  a  remain- 
der. Dividing  each  of  the  numbers  by  2,  the  second  row 
becomes  2,  3,  9,  8,  since  2  will  not  divide  3  and  9  without  a  re- 
mainder. Dividing  again  by  2,  the  result  is  1,  3,  9,  4. 
Dividing  the  third  row  by  3,  the  result  is  1,  1,  3,  4.  So 
continue  until  the  last  row  contains  only  1's.  The  product 
of  all  the  divisors,  or  2x2x3x3x4=  144,  is  the  least 
common  denominator. 

93.     EXAMPLE. — Find  the  least  common  denominator  of  f,  ^,  T7F. 
SOLUTION.—  3  )  9.  12.  18 


2)1.    2,    1 
1,    1,    1 
3X3X2X2  =  36.     Ans. 

94.  To  reduce  two  or   more  fractions  to  frac- 
tions having  a  common  denominator  : 

EXAMPLE. — Reduce  £ ,  f,  and  -£  to  fractions  having  a  common  denomi- 
nator. 

SOLUTION. — The  common  denominator  is  a  number  which  will  con- 
tain 3,  4,  and  2.  The  least  common  denominator  is  12,  because  it  is  the 
smallest  number  which  can  be  divided  by  3,  4,  and  2  without  a  remain- 
der. 

Reducing  f,  3  is  contained  in  12,  4  times.  By  multiplying  both 
numerator  and  denominator  of  £  by  4,  we  find 

O    -vy     A  Q 

•Q  ^  A  =  To-   In  the  same  way  we  find  *  =  TS  and  4  =  A- 
o  X  4       l « 

95.  Rule  8. — Divide  the  common   denominator   by   the 
denominator  of  the  given  fraction,  and  multiply  both  terms  of 
the  fraction  by  the  quotient. 


EXAMPLES  FOR    PRACTICE. 

96.      Reduce  to  fractions  having  a  common  denominator: 
(«)  *,  I  .  4-  f  (a)  |,  I,  i- 

(*)  A,  *>  A-  (6)   A.  II-  A- 

0  4.  A,  If  Ans    I    0   tt-  A,  «• 


(')  A,  A-  A-  (')  if,  A,  W- 

GO  TV,  W,  li-  I  (/)  tt,  H.  It- 


ARITHMETIC.  29 

ADDITION    OF    FRACTIONS. 

97.  Fractions  can  not  be  added  unless  they  have  a  common 
denominator.     We  can  not  add  f  to  £  as  they  now  stand, 
since  the  denominators  represent  parts   of   different  sizes. 
Fourths  can  not  be  added  to  eighths. 

Suppose  we  divide  an  apple  into  4  equal  parts,  and  then 
divide  2  of  these  parts  into  two  equal  parts.  It  is  evident 
that  we  shall  have  2  one-fourths  and  4  one-eighths.  Now  if 
we  add  these  parts  the  result  is  2  +  4  =  0  something.  But 
what  is  this  something?  It  is  not  fourths,  for  six  fourths 
are  1£,  and  we  had  only  1  apple  to  begin  with;  neither  is  it 
eighths,  for  six  eighths  are  f  ,  which  is  less  than  1  apple.  By 
reducing  the  quarters  to  eighths,  we  have  £  =  4»  an(^  adding 
the  other  4  eighths,  4  +  4  =  8  eighths.  The  result  is  cor- 
rect, since  f  =  1.  Or  we  can,  in  this  case,  reduce  the 
eighths  to  quarters.  Thus,  -|  =  f  ;  whence,  adding,  2  +  2  = 
4  quarters,  a  correct  result,  since  £  =  1. 

Before  adding,  fractions  should  be  reduced  to  a  common 
denominator,  preferably  the  least  common  denominator. 

98.  EXAMPLE.—  Find  the  sum  of  i,  f  ,  and  £. 

SOLUTION.  —  The  least  common  denominator  or  the  least  number 
which  will  contain  all  the  denominators  is  8. 

|  =  |,  f  =  f  ,  and  |  =  |. 


EXPLANATION.  —  As  the  denominator  tells  or  indicates  the 
names  of  the/ar/^,  the  numerators  only  are  added  to  obtain 
the  total  number  of  parts  indicated  by  the  denominator. 

99.     EXAMPLE.—  What  is  the  sum  of  12f  ,  14f  ,  and  7^  ? 
SOLUTION.  —  The  least  common  denominator  in  this  case  is  16. 


14*  = 


sum  33  +  f  f  =  33  +  \\\  =  34}J.     Ans. 

The  sum  of  the  fractions  =  f  J  or  1|J,  which  added  to  the  sum  of  the 
whole  numbers  —  b4J. 


30  ARITHMETIC. 

EXAMPLE.— What  is  the  sum  of  17,  13Tsg,  &,  and  3±  ? 
SOLUTION. — The  least  common  denominator  is  32.    13jsT  =  13^*5,  3J  = 
**• 

l*fc 

A 

3A 
^«;«  33ff.     Ans. 

1OO.  Rule  9. — (a)  Reduce  the  given  fractions  to  frac- 
tions having  the  least  common  denominator,  and  write  the 
sum  of  the  numerators  over  the  common  denominator. 

(6)  When  there  are  mixed  mimbers  and  whole  numbers, 
add  the  fractions  first,  and  if  their  sum  is  an  improper  frac- 
tion, reduce  it  to  a  mixed  number  and  add  the  whole  number 
with  the  other  whole  numbers. 


EXAMPLES  FOR  PRACTICE. 
1 0 1 .      Find  the  sum  of 

(«)  f  •  A,  *• 
(*)  t,  A-  «• 
W  i,  I,  A- 

wk*l  Ans^ 

(/)  il  3,  S. 

(<£")   T*T-  A«  H- 
(^    f ,  ii  f- 

1.  The  weights  of  a  number  of  castings  were  412f  lb.,  270^  lb., 
1,020  lb.,  75J  lb.,  and  68£  lb.     What  was  their  total  weight  ? 

Ans.  1,847  lb. 

2.  Four  bolts  are  required,  2|,  If,  2A,  and  l*f  inches  long.     How 
long  a  piece  of  iron  will  be  required  to  cut  them  from,  allowing  f  of  an 
inch  altogether  for  cutting  off  and  finishing  the  ends  ?        Ans.  9^  in. 


(/)!§. 


SUBTRACTION    OF    FRACTIONS. 

1 02,  Fractions  can  not  be  subtracted  without  first  re- 
ducing them  to  a  common  denominator.  This  can  be  shown 
in  the  same  manner  as  in  the  case  of  addition  of  fractions. 

EXAMPLE.— Subtract  f  from  \\. 

SOLUTION. — The  common  denominator  is  16. 


ARITHMETIC.  31 

103.  EXAMPLE. — From  7  take  f. 

SOLUTION. — 1  =  f- ;  therefore,  since  7  =  6  +  1  we  see  that  7  —  6  4-  f , 
so  that  6|  —  f  =  6f .  Ans. 

104.  EXAMPLE.— What  is  the  difference  between  17T\  and  9^f  ? 
SOLUTION. — The  common  denominator  of  the  fractions  is  32.    17T*8  = 

"H- 

minuend  17|| 
subtrahend    9|| 

difference    8^    Ans. 

1 05.  EXAMPLE.— From  9£  take  4T75. 

SOLUTION.— The  common  denominator  of  the  fractions  is  16.     9±  = 

9iV 

minuend  9^       8?§ 

subtrahend  4T7^  or  4^ 
difference  4}f       4£f    Ans. 

EXPLANATION. — As  the  fraction  in  the  subtrahend  is 
greater  than  the  fraction  in  the  minuend,  it  £##  «#/  be  sub- 
tracted; therefore,'  borrow  1,  or  -ff,  from  the  9  in  the  minu- 
end and  add  it  to  the  -ft ;  T\  +  ||  =  ff  -ft  from  f  $  =  ||. 
Since  1  was  borrowed  from  9,  8  remains;  4  from  8  =  4;  4  + 

«  -  ±B- 

1 06.  EXAMPLE.— From  9  take  8T\. 
SOLUTION. —  minuend  9          8^| 

subtrahend  8y\  or  8T\ 

difference    \\         \\     Ans. 

EXPLANATION. — As  there  is  no  fraction  in  the  minuend 
from  which  to  take  the  fraction  in  the  subtrahend,  borrow  1, 
or  ||,  from  9.  ^  from  ||  =  ||.  Since  1  was  borrowed  f  rom 
9,  only  8  is  left.  8  from  8  =  0. 

107.  Rule  1O. — (a)  Reduce  the  fractions  to  fractions 
having  a  common   denominator.      Subtract   one   numerator 
from  the  other  and  place  the  remainder  over  the  common  de- 
nominator. 

(b)  When  there  are  mixed  numbers,  subtract  the  fractions 
and  whole  numbers  separately,  and  place  the  remainders  side 
by  side. 


33  ARITHMETIC. 

(c)  When  the  fraction  in  the  subtraliend  is  greater  than 
the  fraction  in  tlic  minuend,  borrow  1  from  the  whole  number 
in  the  minuend  and  add  it  to  the  fraction  in  the  minuend, 
from  which  subtract  the  fraction  in  the  subtrahend. 

(</)  When  the  minuend  is  a  whole  number,  borrow  1;  re- 
duce it  to  a  fraction  whose  denominator  is  the  same  as  the 
denominator  of  the  fraction  in  the  subtrahend,  and  place  it 
over  that  fraction  for  subtraction. 


EXAMPLES  FOR   PRACTICE. 

1O8.     Subtract 

(a)  iffromH- 

(6)  £  from  if. 

(c)   &  from  &. 

(<0ttfrom«.  Ans. 

(e)  \\  from  |f . 

(/)  13±  from  3% 

(^)  12i  from  27. 

(h)  5^  from  30. 


A- 


(0 


(/)  17*. 
(£0  14f. 
(/&)  24f. 


1.  An  engineer  found  that  he  had  on  hand  48^  gallons  of  cylinder 
oil.     During  the  following  week  he  used  |  of  a  gallon -a  day  for   three 
days,  \  of  a  gallon  on  the  fourth  day,  \\  of  a  gallon  on  the  fifth  day, 
and  \  of  a  gallon  on  the  sixth  day.     How  much  oil  remained  at  the  end 
of  the  week  ?  Ans.  43||  gallons. 

2.  The  main  line  shaft  of  a  manufacturing  plant  is  run  by  an  engine 
and  water  wheel.     A  test  of  the  plant  showed  that  the  engine  was 
capable  of  developing  251f  H.  P.  (horsepower),  and  the  water  wheel, 
under  full   gate,  67^  H.  P.     It   was  also  found   that   the  machinery 
consumed  210j|  H.  P.,  and  the  friction  of  the  shafting  and  belting  was 
32j  H.  P.     How  much  power  remained  unused  ?  Ans.  76f f§  H.  P. 


MULTIPLICATION    OF    FRACTIONS. 

109.  In  multiplication  of  fractions  it  is  not  necessary  to 
reduce  the  fractions  to  fractions  having  a  common  denomi- 
nator. 

110.  Multiplying  the  numerator  or  dividing  the  denom- 
inator multiplies  the  fraction. 


ARITHMETIC.  33 


EXAMPLE.—  Multiply  J  X  4. 
SOLUTION.— 


£  X4=-|X4=  V8  =  3,     Ans 
*X4  =  =  f  =  3.     Ans. 


The  word  "of"  in  multiplication  of  fractions  means  the 
same  as  X  ,  or  times.     Thus, 


£    Of    y^g-  —    ^ 

EXAMPLE.— Multiply  £  by  2. 

SOLUTION.—  £  x  2  =  ^-    *     =  -«-  =  £ ,     Ans. 

or,     |  x  2  =  g-  ^     =  £ .     Ans. 

111.  EXAMPLE.— What  is  the  product  of  T\  and  £  ? 

4X7 
SOLUTION.—         ^  X  i  =  ^g-  ^  -g  =  ^  =  A. 

^  X  7  =       7 
or,  by  cancelation,       y^ _  — —  =  /?.     Ans. 

4 

112.  EXAMPLE.— What  is  |  of  i  of 


113.      EXAMPLE.—  What  is  the  product  of  9f  and  5f  ? 
SOLUTION.—  9f  =  ^  ;  5f  =  \s-. 


114.     EXAMPLE.—  Multiply  15^  by  3. 

SOLUTION.—  15£          15| 

3    or      3 


Ans. 


-  45  +  2f  =  47|.     Ans. 

115.  Rule  11.  —  (a)  Divide  the  product  of  the  numera- 
tors by  the  predict  of  the  denominators.  All  factors  common 
to  the  numerators  and  denominators  should  first  be  cast  out 
by  cancelation. 


34  ARITHMETIC. 

(b\  To  multiply  one  mixed  number  by  another,  reduce  tJion 
both  to  improper  fractions. 

(c)  To  multiply  a  mixed  number  by  a  whole  number,  first 
multiply  the  fractional  part  by  the  multiplier,  and  if  the 
product  is  an  improper  fraction,  reduce  it  to  a  mi. red  number 
and  add  the  whole  number  part  to  the  product  of  the  multi- 
plier and  whole  number. 

EXAMPLES  FOR   PRACTICE. 

116.     Find  the  product  of 

(«)   ?  X  A-  f  («) 

W   14  X  A- 

& 

Ans.  <{  W 

(f)  ''TS» 
(/)  125. 

(A)  if  X  14.  I  (A)   7|. 

1.  A  single  belt  can  transmit •  107J  H.  P.,  but  as  it  is  desired  to  use 
more  power,  a  double  belt  of  the  same  width  is  substituted  for  it. 
Supposing  the  double  belt  to  be  capable  of  transmitting  -^  as  much 
power  as  the   single  belt,  how  many   H.  P.  can  be  used  after  the 
change  ?  Ans.  153^  H.  P. 

2.  What  is  the  weight  of  2|  miles  of  copper  wire  weighing  5} 
pounds  per  100  feet  ?    There  are  5,280  feet  in  a  mile. 

Ans.  796M  pounds. 


3.  The  grate  of  a  steam  boiler  contains  20£  square  feet.  If  the 
boiler  burns  8^  pounds  of  coal  an  hour  per  square  foot  of  grate  area, 
and  can  evaporate  7£  pounds  of  water  an  hour  per  pound  of  coal 
burned,  how  many  pounds  of  water  are  evaporated  by  the  boiler  in 
1  hour  ?  Ans.  1,276|  pounds. 

DIVISION    OF    FRACTIONS. 

117.  In   division   of   fractions   it    is   not   necessary    to 
reduce  the  fractions  to  fractions  having  a  common  denomi- 
nator. 

118.  Dividing  the    numerator  or    multiplying  the   de- 
nominator divides  the  fraction. 


ARITHMETIC.  35 

EXAMPLE.—  Divide  f  by  3. 

SOLUTION.—  When  dividing  the  numerator,  we  have 


=  =  f  =  J.     Ans. 

When  multiplying  the  denominator,  we  have 


O 

SOLUTION.—  f?  -H  2  =  —  ^  g  =  ^.     Ans. 


EXAMPLE.— Divide  ^  by  2. 
SOLUTION. —  Ts?  -H  2  = 

EXAMPLE. — Divide  ^|  by  7. 
SOLUTION.—      it  ^  7  = -33- "^     =  A  =  iV     A°s- 

119.     To  invert  a  fraction  is  to  turn  it  upside  down ; 
that  is,  make  the  numerator  and  denominator  cliange  places. 
Invert  f  and  it  becomes  -| . 

1  2O.      EXAMPLE.—  Divide,^  by  TV 

SOLUTION. — 1.  The  fraction  T\  is  contained  in  T9ff,  3  times,  for  the 
denominators  are  the  same,  and  one  numerator  is  contained  in  the 
other  3  times.  2.  If  we  now  invert  the  divisor,  -fa,  and  multiply, 
the  solution  is 

3 

-., ,.   x\  — 77  —  o.     Ans. 


This  gives  the  j«/«^  quotient  as  in  the  first  case. 

121.      EXAMPLE.—  Divide  £  by  £. 

SOLUTION.—  We  can  not  divide  f  by  ±,  as  in  the  first  case  above,  for 
the  denominators  are  not  the  same  ;  therefore,  we  must  solve  as  in  the 
second  case. 

-       =     X     =  =      or.     An, 


1  22.     EXAMPLE.—  Divide  5  by  \\. 
SOLUTION.  —  \\  inverted  becomes  f§> 


36  ARITHMETIC. 

1  23.      EXAMPLE.—  How  many  times  is  3|  contained  in  7T7ff  ? 
SOLUTION.—  3f  =  ^  ;  7T7ff  =  ^\9. 

^  inverted  becomes  T4g. 


119        4    _  119  X*  _H9_ 
T6-XT5--^X15  ~  W- 


1 24.  Rule    1 2. — Invert  tlie  divisor  and  proceed  as  in 
m  ultiplication. 

1 25.  We  have  learned  that  a  line  placed  between  two 
numbers  indicates  that  the  number  above  the  line  is  to  be 
divided  by  the  number  below  it.     Thus,  -^  shows  that  18  is 
to   be    divided  by  3.     This  is  also  true  if   a  fraction  or  a 
fractional  expression  be  placed  above  or  below  a  line. 

—  means  that  9  is  to  be  divided  by  f  ; means  that 

16 

Q      I      1  \ 

3  X  7  is  to  be  divided  by  the  value  of 


16 


•    is  the  same  as  £  -=-  f . 


It  will  be  noticed  that  there  is  a  heavy  line  between  the  9 
and  the  f .  This  is  necessary,  since  otherwise  there  would 
be  nothing  to  show  whether  9  is  to  be  divided  by  f,  or  -§- 
by  8.  Whenever  a  heavy  line  is  used,  as  in  the  above  case, 
it  indicates  that  all  above  the  line  is  to  be  divided  by  all 
below  it. 

EXAMPLES  FOR  PRACTICE. 
126.     Divide 


(a)  15by6f 

(b)  SObyf. 
to   172  by  f . 

*'  Ans.  4 


(a)  2i. 

(b)  40. 

(c)  215. 

(e)   ios  by  14|  -        (^    ji 6. 

(/)  3£*  by  17f 

(<?")  ii  by  Vj5-- 
(A)  i&  by  723k 

1.     A  j|-inch  boiler  plate  containing  24  square  feet  of  surface  weighs 
362-nr  pounds.     What  is  its  weight  per  square  foot  ?   Ans.  15^  pounds. 


ARITHMETIC.  37 

2.  A  certain  boiler  has  927  1  square  feet  of  heating  surface,  which  is 
equal  to  35  times  the  area  of  the  grate.     What  is  the  area  of  the  grate 
in  square  feet  ?  Ans.  26£  square  feet. 

3.  If  the  distance  around  the  rim  of  a  locomotive  drive  wheel  is 
13^  feet,  how  many  revolutions  will  the  wheel  make  in  traveling  682 
feet  ?  Ans.  52-^  revolutions. 

127.  Whenever  an  expression  like  one  of  the  three 
following  ones  is  obtained,  it  may  always  be  simplified  by 
transposing  the  denominator  from  above  to  below  the  line, 
or  from  below  to  above,  as  the  case  may  be,  taking  care, 
however,  to  indicate  that  the  denominator  when  so  trans- 
ferred is  a  multiplier. 
s  3 

1.  ~  =  -  -  -  =  -fa  =  Jg-  ;  for,  regarding  the  fraction  above 

the  heavy  line  as  the    numerator  of  a   fraction  whose  de- 
nominator is  9,  I  *  4  =  jj-^-j,  as  before. 

2.  -  =  —    —  =  12.     The  proof  is  the  same  as  in  the  first 
I          3 

case. 

5       5x4 

3       -  —  —    —  =  |^  ;   for,    regarding  |  as  the  numerator 
•J      3x9 

8  X  9          5 

of   a  fraction  whose   denominator  is  f,f    '     —  ;  and 

•j*  x  «/     o  /\  &  , 

Qve  — 

"' 


_ 

3X9  —  3  X  9 

4 
This  principle  may  be  used  to  great  advantage  in  cases  like 

X  310  X  f  I  X  72       Reducing  the  mixed  numbers  to  frac- 


40  X  4£  X  5£ 

i  X  310  X  f  J  X  72      AT 
tions,    the    expression     becomes  -  —  -.  —  .     Mow 

transferring  the  denominators  of  the  fractions  and  canceling, 

;0        3        0  3 

1  X  310  X  27  X  72  X  2  x  6  _  1  X  CT  X  2/  X  flZ  X  ?  X  0  = 
40  X  9  X  31  X  4  X  12  ^0  X  ^  X  ^  X  £  X  }$ 

i  2 


38  ARITHMETIC. 

Greater  exactness  in  results  can  usually  be  obtained  by 
using  this  principle  than  can  be  obtained  by  reducing  the 
fractions  to  decimals.  The  principle,  however,  should  not 
be  employed  if  a  sign  of  addition  or  subtraction  occurs  either 
above  or  below  the  dividing  line. 


DECIMALS. 

1 28.  Decimals  are  tentJi  fractions  ;  that  is,  the  parts 
of  a  unit  are  expressed  on  the  scale  of  ten,  as  tenths,  hun- 
dredths,  thousandths,  etc. 

1 29.  The  denominator,  which  is  always  ten  or  a  multiple 
of  ten,  as  10,  100,  1,000,  etc.,  is  not  expressed  as  it  would  be 
in  common   fractions,   by  writing  it  under  the  numerator, 
with  a  line  between  them;  as,   -^,  yf^,    ^ 030 0 .     The  denomi- 
nator is  always  understood,  the  numerator  consisting  of  the 
figures  on  the  right  of  the  unit  figure.     In  order  to  distinguish 
the  unit  figure,  a  period  (.),  called  the  decimal  point, 
is  placed  between  the  unit  figure  and  the  next  figure  on  the 
right.     The  decimal  point  may  be  regarded  in  two  ways  : 
first,  as  indicating  that  the  number  on  the  right  is  the  nu- 
merator of  a  fraction  whose  denominator  is  10,  100,  1,000,  etc. ; 
and,  second,  as  a  part  of  the  Arabic  system  of  notation, 
each  figure  on  the  right  being  10  times  as  large  as  the  next 
succeeding  figure,  and  10  times  as  small  as  the  next  prece- 
ding figure,  serving  merely  to  point  out  the  unit  figure. 

130.  The  reading  of  a  decimal  number  depends  upon  the 
number  of  decimal  places  in  it,  or  the  number  of  figures  to 
the  right  of  the  unit  figure. 

The  first  figure  to  the  right  of  the  unit  figure  expresses 
tenths. 

The  second  figure  to  the  right  of  the  unit  figure  expresses 
hundredths. 

The  third  figure  to  the  right  of  the  unit  figure  expresses 
thousandths. 

The  fourth  figure  to  the  right  of  the  unit  figure  expresses 
ten  -  thousa  ndths. 

The  fifth  figure  to  the  right  of  the  unit  figure  expresses 
hundred-thousandths. 


ARITHMETIC.  39 

The  sixth  figure  to  the  right  of  the  unit  figure  expresses 
millionths. 
Thus: 

.3  fV       =  3  tenths. 

.03          =      yfjj-      =3  hundredths. 
.003  T^     =  3  thousandths. 

•0003  ^Ihnj-  =  3  ten-thousandths. 

.00003    ==  ronroo  =  3  hundred-thousandths. 
.000003  =  tooilooo  =  3  millionths. 

The  first  figure  to  the  right  of  the  unit  figure  is  called  the 
first  decimal  place  ;  the  second  figure,  the  second  decimal 
place,  etc.  We  see  in  the  above  that  the  number  of  decimal 
places  in  a  decimal  equals  the  number  of  cipJiers  to  the  right 
of  the  figure  1  in  the  denominator  of  its  equivalent  fraction. 
This  fact  kept  in  mind  will  be  of  much  assistance  in  reading 
and  writing  decimals. 

Whatever  may  be  written  to  the  left  of  a  decimal  point  is 
a  whole  number.  The  decimal  point  affects  only  the  figures 
to  its  riglit. 

When  a  wJiole  number  and  decimal  are  written  together, 
the  expression  is  a  mixed  number.  Thus,  8.12  and  17. 25  are 
mixed  numbers. 

The  relation  of  decimals  and  whole  numbers  to  each  other 
is  clearly  shown  by  the  following  table  : 


millions, 
ons. 

thousands. 

1 

rt 
en 

Iths. 

usandths. 

,1 

<_i_i   **"' 

, 

fl 

c 

i/i 

3      0 

""5  '~~i 

hundreds  o: 
tens  of  mil 

millions. 

hundreds  o 

tens  of  tho 
thousands. 

hundreds. 

t 

u 

units. 

decimal  poi 
tenths. 

hundredths 
thousandth 

ten-thousar 
hundred-th 

millionths. 

ten-millioni 
hundred-m 

9     8 

7 

6 

5  4 

3 

% 

1 

.     2 

3    4 

5    6 

7 

8    9 

The  figures  to  the  left  of  \.\\Q  decimal  point  represent  whole 
numbers  ;  those  to  the  rig/it  are  decimals. 

In  both  the  decimals  and  whole  numbers,  the  units  place 
is  made  the  starting  point  of  notation  and  numeration.  The 


40  ARITHMETIC. 

decimals  decrease  on  the  scale  of  ten  to  the  right  and 
the  zvhole  numbers  increase  on  the  scale  of  ten  to  the  left. 
The  first  figure  to  the  left  of  units  is  tens,  and  the  first  figure 
to  the  rig/it  of  units  is  tenths.  The  second  figure  to  the  left  of 
units  is  hundreds,  and  the  second  figure  to  the  right  is  hun- 
dredths.  The  third  figure  to  the  left  is  thousands,  and  the 
thirds  the  rz^v^  is  thousandths,  and  so  on  ;  the  whole  num- 
bers on  the  left  and  the  decimals  on  the  right.  The  figures 
equally  distant  from  units  place  correspond  in  name.  The 
decimals  have  the  ending  tits,  which  distinguishes  them  from 
whole  numbers.  The  following  is  the  numeration  of  the 
number  in  the  above  table  :  Nine  hundred  eighty-seven 
million,  six  hundred  fifty-four  thousand,  three  hundred 
twenty-one,  and  twenty-three  million,  four  hundred  fifty- 
six  thousand,  seven  hundred  eighty-nine  hundred  millionths. 
The  decimals  increase  to  the  left  on  a  scale  of  ten,  the 
same  as  whole  numbers,  for  if  you  begin  at  ^-thousandths, 
in  the  table,  you  see  that  the  next  figure  is  hundredths, 
which  is  ten  times  as  great,  and  the  next  tenths,  or  ten  times 
the  hundredths,  and  so  on  through  both  decimals  and  whole 
numbers. 

131.  Annexing  or  taking  away  a  cipher  at  the  right  of 
a  decimal  does  not  affect  its  value. 

.5  is  fV;  -50  is  tV°o,  but  T5¥  =  AV;  therefore,  .5  =  .50. 

132.  Inserting  a   cipher   between   a   decimal    and   the 
decimal  point  divides  the  decimal  by  10. 

•S  =  A;  fV-10  =  T!«r=.05. 

1 33.  Taking  away  a  cipher  from  the  left  of  a  decimal 
multiplies  the  decimal  by  10. 

•05  =         ;          x  10  =       **•*• 


ADDITION    OF    DECIMALS. 

134.  The  only  respect  in  which  addition  of  decimals 
differs  from  addition  of  whole  numbers  is  in  the  placing  of 
the  numbers  to  be  added. 

Whole  numbers  begin  at  units  and  increase  on  the  scale 
of  10  to  the  left.  Decimals  decrease  on  the  scale  of  10  to 


ARITHMETIC.  41 

the  right.  Whole  numbers  are  to  the  left  of  the  decimal 
point  and  decimals  are  to  the  right  of  it.  In  whole  numbers 
the  right  hand  side  of  a  column  of  figures  to  be  added  must 
be  in  line,  and  in  decimals  the  left  hand  side  must  be  in 
line,  which  brings  the  decimal  points  directly  under  each 
other. 

whole  numbers  decimals  mixed  numbers 

342  .342  342.032 

4234  .4234  4234.5 

26  .26  26.6782 

3  .03  3.06 


sum  4605  Ans.         sum  1.0554  Ans.          sum  4606.2702  Ans. 

135.      EXAMPLE.— What  is  the  sum  of  242,  .36,  118.725,  1.005,  6, 
and  100.1? 

SOLUTION. —  242. 

.36 

118.725 
1.005 
6. 
100.1 


sum  468.190    Ans. 

136.  Rule  13. — Place  the  numbers  to  be  added  so  tJiat 
the  decimal  points  will  be  directly  under  each  other.  Add  as 
in  whole  numbers,  and  place  the  decimal  point  in  the  sum 
directly  under  the  decimal  points  above. 


EXAMPLES  FOR   PRACTICE. 

1  37.  Find  the  sum  of 
(a)  .2143,  .105,  2.3042,  and  1.1417. 
(V)  783.5,  21.473,  .2101,  and  .7816. 
(c)  21.781,  138.72,  41.8738,  .72,  and  1.413. 
(</)  .3724,  104.15,  21.417,  and  100.042.         A 
(e)  200.172,  14.105,  12.1465,  .705,  and  7.2. 
(/)  1,427.16,  .244,  .32,  .032,  and  10.0041. 
(g)  2,473.1,  41.65,  .7243,  104.067,  and  21.073. 
(K)  4,107.2,  .00375,  21.716,  410.072,  and  .0345. 


(a)  3.7652. 

(b)  805.9647. 
(f)  204.5078. 
(//)  225.9814. 
(e)   234.3285. 
(/)  1,437.7601. 
(g)  2,640.6143. 
(//)  4,539.02625. 


1.  The  estimated  weights  of  the  parts  of  a  return  tubular  boiler 
were  as  follows:  Shell,  3,626  Ib. ;  tubes,  3,564.5  Ib. ;  manhole  cover, 
ring,  and  yoke,  270.34  Ib. ;  stays,  etc.,  1,089.4  Ib. ;  steam  nozzles, 


42  ARITHMETIC. 

236.07  Ib. ;  handhole  covers  and  yokes,  120.25  Ib. ;  feed  pipe,  34.75  Ib. ; 
boiler  supports,  350.6  Ib.  What  was  the  total  estimated  weight  of  the 
boiler?  Ans.  9,291.91  Ib. 

2.  A  bill  for  engine  room  supplies  had  the  following  items:  1  waste 
can,  $8.30;  20  feet  of  4-inch  belting,  $11.20;  1  pipe  wrench,  $1.65;  12 
pounds  of  waste,  $0.84;  5  gallons  of  cylinder  oil,  $8.75;  20  gallons  of 
machine  oil,  $24.  How  much  did  the  bill  amount  to  ?  Ans.  $54.74. 


SUBTRACTION    OF    DECIMALS. 

138.  For  the  same  reason  as  in  addition  of  decimals, 
the  left  hand  figures  of  decimal  numbers  are  placed  in  line 
and  the  decimal  points  under  each  other. 

EXAMPLE.— Subtract  .132  from  .3063. 
SOLUTION.  —  minuend  .3063 

subtrahend  .132 

difference  .1  743    Ans. 

1 39.  EXAMPLE. — What  is  the  difference  between  7.895  and  .725  ? 
SOLUTION. —  minuend  7.895 

subtrahend    .725 


difference  7.170  or  7. 17  Ans. 

1 4O.      EXAMPLE.— Subtract  .625  from  11. 
SOLUTION. —  minuend  1 1.000 

subtrahend       .625 


difference  10.375 

141.  Rule  14. — Place  the  subtrahend  tinder  the  minu- 
end, so  that  the  decimal  points  will  be  directly  under  each 
other.  Subtract  as  in  wJiole  numbers,  and  place  the  decimal 
point  in  the  remainder,  directly  under  the  decimal  points  above. 

When  the  figures  in  the  decimal  part  of  the  subtrahend 
extend  beyond  those  in  the  minuend,  place  ciphers  in  the 
minuend  above  them  and  subtract  as  before. 


EXAMPLES  FOR   PRACTICE. 
142.      From 

( a )  407. 385  take  235. 0004.  f  ( a )  172. 3846. 

(6)  22.718  take  1.7042.  A         I  '  (g)  21.0138. 

(c)   1,368. 17  take  13. 6817.  S'  1    (f)   1,354.4883. 

(tf)  70.00017  take  7.000017.  [  (</)  63.000153. 


ARITHMETIC.  43 


(e)   630.630  take  .6304.  [    (e) 

(/)  421.73  take  217.162.  Ang     I  (/)  204.568. 
(^•)  1.000014  take  .00001.  1  (^)  1.000004. 

(A)  .783652  take  .542314.  [  (//)   .241338. 

1.  If  the  temperature  of  steam  at  5  pounds  pressure   is  227.964 
degrees,  and  at  100  pounds  pressure  337.874  degrees,  how  many  degrees 
warmer  is  the  steam  at  the  higher  pressure  ?  Ans.  109.91  degrees. 

2.  The    outside   diameter    of  2i-inch    wrought-iron    pipe    is    2.87 
inches  and  the  inside  diameter  2.46  inches.     How  thick  is  the  pipe  ? 

Ans.  .41  H-  2  =  .205  inch. 

3.  In  a  cistern  that  will  hold  326.5  barrels  of  water  there  are  178.625 
barrels.     How  much  does  it  lack  of  being  full  ?      Ans.   147.875  barrels. 

4.  A  wrought-iron  rod  is  2.53  inches  in  diameter.     What  must  be 
the  thickness  of  metal  turned  off  so  that  the  rod  will  be  2.495  inches  in 
diameter?  Ans.  .035  --  2  =  .0175. 


MULTIPLICATION    OF    DECIMALS. 

143.  In  multiplication  of  decimals,  we  do  not  place  the 
decimal  points  directly  under  each  other  as  in  addition  and 
subtraction.  We  pay  no  attention  for  the  time  being  to  the 
decimal  points.  Place  the  multiplier  under  the  multiplicand, 
so  that  the  right-hand  figure  of  the  one  is  under  the  right- 
hand  figure  of  the  other,  and  proceed  exactly  as  in  multipli- 
cation of  whole  numbers.  After  multiplying,  count  t/i£ 
number  of  decimal  places  in  botli  multiplicand  and  multiplier, 
and  point  off 'the  same  number  in  the  product. 

EXAMPLE.— Multiply  .825  by  13. 
SOLUTION.—        multiplicand       .825 
multiplier  1 3 

2475 
825 


product    10.725    Ans. 

In  this  example  there  are  three  decimal  places  in  the  mul- 
tiplicand and  none  in  the  multiplier;  therefore,  3  decimal 
places  are  pointed  off  in  the  product. 


44  ARITHMETIC. 

1 44.  EXAMPLE.  —What  is  the  product  of  426  and  the  decimal .  005  ? 
SOLUTION.—          multiplicand       426 

multiplier      .005 

product    2.1  30  or  2. 13    Ans. 

In  this  example  there  are  3  decimal  places  in  the  multi- 
plier and  none  in  the  multiplicand;  therefore,  3  decimal 
places  are  pointed  off  in  the  product. 

145.  It  is  not  necessary  to  multiply  by  the  ciphers  on  the 
left  of  a  decimal;  they  merely  determine  the  number  of  deci- 
mal places.     Ciphers  to  the  right  of  a  decimal  should  be  re- 
moved, as  they  only  make  more  figures  to  deal  with,  and  do 
not  change  the  value. 

1 46.  EXAMPLE.— Multiply  1.205  by  1.15. 
SOLUTION.—      multiplicand         1.205 

multiplier  1.15 

6025 
1205 
1  205 


product    1.38575    Ans. 

In  this  example  there  are  3  decimal  places  in  the  multi- 
plicand, and  2  in  the  multiplier,  therefore  3  -}-  2,  or  5,  deci- 
mal places  must  be  pointed  off  in  the  product. 

147.  EXAMPLE.— Multiply  .232  by  .001. 
SOLUTION. —         multiplicand          .232 

multiplier          .001 

product  .000232    Ans. 

In  this  example  we  multiply  the  multiplicand  by  the  digit 
in  the  multiplier,  which  makes  232  in  the  product,  but  since 
there  are  3  decimal  places  in  each,  the  multiplier  and  the 
multiplicand,  we  must  prefix  3  ciphers  to  the  232,  to  make 
3  +  3,  or  6,  decimal  places  in  the  product. 

148.  Rule  15. — Place  the  multiplier  under  the  multipli- 
cand, disregarding  the  position  of  the  decimal  points.     Mul- 
tiply as  in  whole  numbers,  and  in  the  product  point  off  as 


ARITHMETIC.  45 

many  decimal  places  as  there  are  decimal  places  in  both  mul- 
tiplier and  multiplicand,  prefixing  ciphers  if  necessary. 


EXAMPLES  FOR   PRACTICE. 

1 49.      Find  the  product  of 


(a)  .000492x4.1418. 

(6)  4,003.2x1.2. 

(c)  78.6531  x  1-03. 

(</)  .3685  X-  042.  . 

(e)  178,352  X. 01. 

(/)  .00045  X  .0045. 

(£•)  .714  X  .00002. 

(k)  .00004  X  -008. 


(a)  .0020377656. 
(6)  4,803.84. 

(c)  81.012693. 

(d)  .015477. 
(<?)  1,783.52. 
(/)  .000002025. 
(g)  .00001428. 
(h)  .00000032. 


1.  The  stroke  of  an  engine  was  found  by  measurement  to  be  2.987 
feet.     How  many  feet  will  the  cross-head  pass  over  in  600  revolutions  ? 

Ans.    3,5844  feet. 

2.  If  a  steam  pump  delivers  2.39  gallons  of  water  per  stroke  and 
runs  at  51  strokes  a  minute,  how  many  gallons  of  water  would  it  pump 
in  58£  minutes  1  Ans.  7,130.565  gallons. 

3.  Wishing    to    obtain    the   weight    of   a   connecting-rod  from  a 
drawing,  it  was  calculated  that  the  rod  contained  294.8  cubic  inches  of 
wrought  iron,  63.5  cubic  inches  of  brass,  and   10.4  cubic   inches  of 
babbitt.     Assuming  the  weight  of  wrought  iron  to  be  .278  pound  per 
cubic  inch,  of  brass  .303  pound,  and  of  babbitt  .264  pound,  what  was  the 
weight  of  the  rod  7  Ans.  103.94  pounds. 


DIVISION    OF    DECIMALS. 

15O.  In  division  of  decimals  we  pay  no  attention  to  the 
decimal  point  until  after  the  division  is  performed.  The 
number  of  decimal  places  in  the  dividend  must  equal  (be  made 
to  equal  by  annexing  ciphers]  the  number  of  decimal  places  in 
the  divisor.  Divide  exactly  as  in  whole  numbers.  Subtract 
the  number  of  decimal  places  in  the  divisor  from  the  number  of 
decimal  places  in  the  dividend,  and.  point  off 'as  many  decimal 


46  ARITHMETIC. 

places  in  the  quotient  as  there  are  units  in  the  remainder  thus 
found. 

EXAMPLE.— Divide  .625  by  25. 

divisor  dividend  quotient 

SOLUTION.—  25).625(.025    Ans. 

50 

125 
125 

remainder       0 

In  this  example  there  are  no  decimal  places  in  the  divisor, 
and  3  decimal  places  in  the  dividend;  therefore,  there  are  3 
minus  0,  or  3,  decimal  places  in  the  quotient.  One  cipher 
has  to  be  prefixed  to  the  25,  to  make  the  3  decimal  places. 

151.     EXAMPLE.— Divide  6.035  by  .05. 

divisor    dividend    quotient 

SOLUTION.—  .0  5  )  6.0  3  5  (  1  2  0.7    Ans. 

5 

To 

10 

35 
35 

remainder    0 

In  this  example  we  divide  by  5,  as  if  the  cipher  were  not 
before  it.  There  is  one  more  decimal  place  in  the  dividend 
than  in  the  divisor ;  therefore,  one  decimal  place  is  pointed 
off  in  the  quotient. 

1  52.      EXAMPLE.— Divide  .125  by  .005. 

divisor  dividend  quotient 

SOLUTION.—  .0  0  5  )  .1  2  5  (  2  5    Ans. 

1  0 

~25 
25 

remainder    0 

In  this  example  there  are  the  same  number  of  decimal 
places  in  the  dividend  as  in  the  divisor ;  therefore,  the  quo- 
tient has  no  decimal  places,  and  is  a  whole  number. 


ARITHMETIC.  47 

1 53.      EXAMPLE.— Divide  326  by  .25. 

divisor     dividend    quotient 

SOLUTION. —  .2  5  )  3  2  6.0  0  (  1  3  0  4    Ans. 

25 

76 

75 


100 
1  00 

remainder      0 

In  this  problem  two  ciphers  were  annexed  to  the  dividend, 
to  make  the  number  of  decimal  places  equal  to  the  number 
in  the  divisor.  The  quotient  is  a  whole  number. 

1 54.  EXAMPLE.— Divide  .0025  by  1.25. 
SOLUTION.—  1.2  5  )  .0  0  2  5  0  (  .002    Ans. 

250 

remainder      0 

EXPLANATION. — In  this  example  we  are  to  divide  .0025  by 
1.25.  Consider  the  dividend  as  a  whole  number,  or  25  (dis- 
regarding the  two  ciphers  at  its  left,  for  the  present) ;  also, 
consider  the  divisor  as  a  whole  number,  or  125.  It  is  clearly 
evident  that  the  dividend  25  will  not  contain  the  divisor  125 ; 
we  must,  therefore,  annex  one  cipher  to  the  25,  thus  making 
the  dividend  250.  125  is  contained  twice  in  250,  so  we  place 
the  figure  2  in  the  quotient.  In  pointing  off  the  decimal 
places  in  the  quotient,  it  must  be  remembered  that  there 
were  only  four  decimal  places  in  the  dividend;  but  one 
cipher  was  annexed,  thereby  making  4+1,  or  5,  decimal 
places.  Since  there  are  5  decimal  places  in  the  dividend  and 
2  decimal  places  in  the  divisor,  we  must  point  off  5  —  2,  or  3, 
decimal  places  in  the  quotient.  In  order  to  point  off  3  deci- 
mal places,  two  ciphers  must  be  prefixed  to  the  figure  2, 
thereby  making  .002  the  quotient.  It  is  not  necessary  to 
consider  the  ciphers  at  the  left  of  a  decimal  when  dividing, 
except  when  determining  the  position  of  the  decimal  point  in 
the  quotient. 

155.  Rule  16. — (a)  Place  the  divisor  to  the  left  of  t/ie 
dividend  and  proceed  as  in  division  of  whole  numbers,  and  in  the 


48  ARITHMETIC. 

quotient,  point  off  as  many  decimal  places  as  the  number  of 
decimal  places  in  the  dividend  exceed  those  in  the  divisor,  pre- 
fixing ciphers  to  the  quotient,  if  necessary. 

(b)  If  in  dividing  one  number  by  another  there  be  a  re- 
mainder, the  remainder  can  be  placed  over  the  divisor,  as  a 
fractional  part  of  the  quotient,  but  it  is  generally  better  to 
annex  ciphers  to  the  remainder,  and  continue  dividing  until 
there  are  3  or  4  decimal  places  in  the  quotient,  and  then  if 
there  still  be  a  remainder,  terminate  the  quotient  by  the  plus 
sign  (+).  which  shows  that  it  can  be  carried  further. 

1 56.     EXAMPLE.— What  is  the  quotient  of  199  divided  by  15  ? 

SOLUTION.—  15)199(13  +  ^    Ans. 

15 

49 
45 

remainder      4 

Or,       15)199.000(13.266+    Ans. 
15 

T9 
45 

~40 
30 

f(M) 
90 
100 
90 

remainder      1 0 
13^  =  13.266  + 
A=     .266  + 

It  very  frequently  happens  as  in  the  above  example, 
that  the  division  will  never  terminate.  In  such  cases,  de- 
cide to  how  many  decimal  places  the  division  is  to  be  carried, 
and  carry  the  work  one  place  further.  If  the  last  figure  of 
the  quotient  thus  obtained  is  5  or  a  greater  number,  increase 
the  preceding  figure  by  1,  and  write  after  it  the  minus  sign 
(  — ),  thus  indicating  that  the  quotient  is  not  quite  as  large 
as  indicated;  if  the  figure  thus  obtained  is  less  than  5,  write 
the  plus  sign  (-)-)  after  the  quotient,  thus  indicating  that 


ARITHMETIC.  49 

the  number  is  slightly  greater  than  as  indicated.  In  the  last 
example,  had  it  been  desired  to  obtain  the  answer  correct  to 
four  decimal  places,  the  work  would  have  been  carried  to 
five  places,  obtaining  13.26666,  and  the  answer  would  have 
been  given  as  13.2667  —  .  This  remark  applies  to  any  other 
calculation  involving  decimals,  when  it  is  desired  to  omit 
some  of  the  figures  in  the  decimal.  Thus,  if  it  is  desired  to 
retain  three  decimal  places  in  the  number  .2471253,  it  would 
be  expressed  as  .247+  ;  if  it  was  desired  to  retain  five 
decimal  places,  it  would  be  expressed  as  .24713  — .  Both  the 
-j-  and  —  signs  are  frequently  omitted  ;  they  are  seldom 
used  outside  of  arithmetic,  except  in  exact  calculations, 
when  it  is  desired  to  call  particular  attention  to  the  fact 
that  the  result  obtained  is  not  quite  exact. 


EXAMPLES  FOR  PRACTICE. 

157.      Divide 

(a)  101. 6688  by  2.36. 
(£)   187. 12264  by  123.107. 


(c)   .08  by  .008. 
(</)  .0003  by  3.75. 
0?)   .0144  by  .024 
(/)  .00375  by  1.25. 
(g)  .004  by  400. 


Ans.  - 


(a)  43.08. 
(6)  1.52. 
(c)   10. 

(d)  .00008. 

(e)  .6. 
(/)  .003. 
(g)  .00001. 
(/;)  50. 


(h)  .4  by  .008. 

1.  In  a  steam  engine  test  of  an  hour's  duration,  the  horsepower 
developed  was  found  to  be  as  follows,  at  10-minute  intervals  :   25.73, 
25.64,   26.13,  25.08,  24.20,  26.7,  26.34.     What  was  the  average  horse- 
power? Ans.  25.6886,  average. 

NOTE. — Add  the  different  horsepowers  together  and  divide  by  the 
number  of  tests,  or  7. 

2.  There  are  31.5  gallons  in  a  barrel.     How  many  barrels  are  there 
in  2,787.75  gallons  1  Ans.  88.5  barrels. 

3.  A  car  load  of  18.75  tons  of  coal  cost  $60.75.     How  much  was  it 
worth  per  ton  7  Ans.  $3.24  per  ton. 

4.  A  keg  of  TV  by  If-inch  boiler  rivets  weighs  100  pounds  and 
contains  595  rivets.     What  is  the  weight  of  one  of  the  rivets  1 

Ans.  .168  pound. 


50  ARITHMETIC. 


TO  REDUCE  A  FRACTION  TO  A  DECIMAL. 

158.      EXAMPLE.  —  f  equals  what  decimal  ? 

SOLUTION.—  4)3.00 

—  —  —  ,  or  £  =  .75.     Ans. 

•  7  u 

EXAMPLE.  —  What  decimal  is  equal  to  £  ? 
SOLUTION.—  8)7.000(.875 

64 


56        or|  =  .875.     Ans. 

To" 

40 


159.  Rule  17.  —  Annex  ciphers  to  the  numerator  and 
divide  by  the  denominator.  Point  off  as  many  decimal 
places  in  the  quotient  as  there  are  ciphers  annexed. 


EXAMPLES  FOR   PRACTICE. 

16O.      Reduce  the  following  common  fractions  to  decimals: 


(«)  if- 

(b)  |. 

(f)  H- 

(d)  ft. 

W  A- 


Ans.  * 


(a)  .46875. 

(b)  .875. 
(<r)   .65625. 

(d)  .796875. 

(e)  .16. 
(/)  .625. 
(g)  .05. 

>J    .004. 


161.  To  reduce  inches  to  decimal  parts  of  a 
foot  : 

EXAMPLE.— What  decimal  part  of  a  foot  is  9  inches  ? 

SOLUTION.  — Since  there  are  12  inches-  in  one  foot,  1  inch  is  ^  of  a 
foot,  and  9  inches  is  9  x  r?  or  r'?  of  a  foot.  This  reduced  to  a  decimal 
by  the  above  rule,  shows  what  decimal  part  of  a  foot  9  inches  is. 

12)9.00(.75ofafoot.     Ans. 
84 


60 

b~ 


ARITHMETIC.  51 

162.  '  Rule  18. — (a)  To  reduce  inches  to  decimal  parts 
of  a  foot,  divide  the  number  of  inches  by  12. 

(b)  Should  the  resulting  decimal  be  an  unending  one  and  it 
is  desired  to  terminate  the  division  at  some  point,  say  the 
fourth  decimal  place,  carry  the  division  one  place  fartJier^ 
and  if  the  fifth  figure  is  5  or  greater,  increase  the  fourth 
figure  by  one  and  omit  the  sign  -}-. 


EXAMPLES   FOR   PRACTICE. 

163.      Reduce  to  the  decimal  part  of  a  foot: 
(a)  3  in. 
(6)  4|.  in. 
(c)    5  in.  Ans. 

(d)  6f  in. 

(e)  11  in. 

1.  The  lengths  of  belting  required  to  connect  three  countershafts 
with  the   main   line  shaft  were   found   with  a  tape  measure   to  be 
27  ft.  4  in.,  23  ft.  8  in.,  and  38  ft.  6  in.     How  many  feet  of  belting  were 
necessary?  Ans.  89.5ft. 

2.  The  stroke  of  an  engine  is  14  inches.     What  is  the  length  of  the 
crank  measured  from  the  center  of  shaft  to  center  of  crank-pin  in  feet  ? 

Ans.  .5833-h  foot. 

3.  A  steam  pipe  fitted  with  an  expansion  joint  was  found  to  expand 
1.668  inches  when   steam   was  admitted   to  it.     How   much   was  its 
expansion  in  decimal  parts  of  a  foot?  Ans.  .139  foot. 


TO  REDUCE  A  DECIMAL  TO  A  FRACTION. 

164.  EXAMPLE.— Reduce  .125  to  a  fraction. 
SOLUTION.— .  125  =  -j1^  =  T65  =  |.     Ans. 

EXAMPLE.— Reduce  .875  to  a  fraction. 
SOLUTION.— .875  =  TV555  =  f  $  =  f .     Ans. 

1 65.  Rule   1 9. —  Under  the  figures  of  the  decimal,  place 
the  digit  1  with  as  many  ciphers  at  its  right  as  there  are 
decimal  places  in  the  decimal,  and  reduce  the  resulting  fraction 
to  its  lowest  terms  by  dividing  both  numerator  and  denomi- 
nator by  the  same  number. 


52 


ARITHMETIC. 


166. 


EXAMPLES  FOR    PRACTICE. 

Reduce  the  following  to  common  fractions: 


(a)  .125. 

(b)  .625. 

(c)  .3125. 

(d)  .04. 

(e)  .06. 


Ans.  - 


(/)  -75. 
(g)  .15625. 
(k)  .875. 


(«)  i- 
(*)  f- 
(0  A- 

(<*)  A- 
(0   A- 
</>*• 
(£•)  A- 
(4)   i- 


TO  EXPRESS  A  DECIMAL  APPROXIMATELY 
AS  A  FRACTION  HAVING  A  GIVEN 

DENOMINATOR. 
167.     EXAMPLE.— Express  .5827  in  64ths. 
SOLUTION. -.5827  x  ft  =  87'^28.  say  |f 
Hence,  .5827  =  |J,  nearly.     Ans. 

EXAMPLE.— Express  .3917  in  12ths. 
A  7004. 

SOLUTION.— .3917  x  \\  =  ^j^.  say  A- 

Hence,  .3917  =  TB2,  nearly.     Ans. 

1  68.  Rule  2O. — Reduce  1  to  a  fraction  having  the  given 
denominator.  Multiply  the  given  decimal  by  tiic  fraction  so 
obtained,  and  the  result  will  be  the  fraction  required. 


EXAMPLES   FOR    PRACTICE. 
1 69.      Express 

(a)  .625  in  8ths. 

(b)  .3125  in  16ths. 

(c)  .15625  in  32ds. 

(d)  .77  in  64ths. 

(e)  .81  in  48ths. 

(/)  .923  in  96ths.  (/)  gf. 


Ans. 


A- 
A- 


ARITHMETIC.  53 

1  7O.  The  sign  for  dollars  is  $.  It  is  read  dollars.  $25 
is  read  25  dollars. 

Since  there  are  100  cents  in  a  dollar,  one  cent  is  1-one- 
hundredth  of  a  dollar;  the  .first  two  figures  of  a  decimal 
part  of  a  dollar  represent  cents.  Since  a  mill  is  ^  of  a  cent, 
or  -ruVfi-  of  a  dollar,  the  third  figure  represents  mills. 

Thus,  $25.16  is  read  twenty-five  dollars  and  sixteen  cents; 
$25.168  is  read  twenty-five  dollars,  sixteen  cents  and  eight 
mills. 

171.  The   vine  11 1 u in  ,  parenthesis  (  ),   bracket 

[  ],   and  brace   {  }  are  called  symbols  of  aggregation, 

and  are  used  to  include  numbers  which  are  to  be  considered 
together;  thus,  13  X  8-3,  or  13  X  (8  —  3),  shows  that  3  is  to 
be  taken  from  8  before  multiplying  by  13. 

13  X  (8  —  3)  =  13  X  5  =  65.     Ans. 

13  X    8-3   =  13  X  5  =  65.     Ans. 
When  the  vinculum  or  parenthesis  is  not  used,  we  have 

13  X  8  -  3  =  104  —  3  =  101.     Ans. 

1 72.  In  any  series  of  numbers  connected  by  the  signs  -|-, 
— ,  X,  and  -+,  the  operations  indicated  by  the  signs  must  be 
performed  in  order  from  left  to  right,  except  that  no  addition 
or  subtraction  may  be  performed  if  a  sign  of  multiplication 
or  division  follow?  the  number  on  the  right  of  a  sign  of 
addition  or  subtraction,  until  the  indicated  multiplication  or 
division  has  been  performed.     In  all  cases  the  sign  of  multi- 
plication takes  the  precedence,  the  reason  being  that  when 
two  or  more  numbers  or  expressions  are  connected  by  the 
sign  of  multiplication,  the  numbers  thus  connected  are  re- 
garded as  factors  of  the  product  indicated,  and  not  as  sepa- 
rate numbers. 

EXAMPLE.— What  is  the  value  of  4  x  24  —  8  4- 17  ? 
SOLUTION.— Performing  the  operations  in  order  from  left  to  right, 
4x24  =  96;  96-8  =  88;  88  +  17  =  105.     Ans. 

173.  EXAMPLE.— What  is  the  value  of  the  following  expression  : 
1,296  -i- 12  +  160  -  22  X  3£  =  ? 

1  SOLUTION.— 1,296-H  12  =  108;  108  +  160  =  268;  here  we  cannot  sub- 
tract 22  from  268  because  the  sign  of  multiplication  follows  22;  hence, 
multiplying  22  by  3$,  we  get  77,  and  268  -  77  =  191.  Ans. 


54  ARITHMETIC. 

Had  the  above  expression  been  written  1,296  ~  12  +  160  — 
22  X  3|-  -4-7  +  25,  it  would  have  been  necessary  to  have  divi- 
ded 22  X  3|-  by  7  before  subtracting,  and  the  final  result  would 
have  been  22  X  3$  =  77;  77  -J-.7  =  11;  268  -  11  =  257;  257 
-j-  25  =  282.  Ans.  In  other  words,  it  is  necessary  to  per- 
form all  of  the  multiplication  or  division  included  between 
the  signs  -{-  and  — ,  or  —  and  -{-,  before  adding  or  subtracting. 
Also,  had  the  expression  been  written  1,290  -4-  12  +  160 
—  24£  -4-  7  X  3£  +  25,  it  would  have  been  necessary  to  have 
multiplied  3^-  by  7  before  dividing  24£,  since  the  sign  of 
multiplication  takes  the  precedence,  and  the  final  result 
would  have  been  3£  X  7  =  24£;  24£  -4-  24£  ==  1;  268  -  1  = 
267 ;  267  +  25  —  292.  Ans. 

It  likewise  follows  that  if  a  succession  of  multiplication 
and  division  signs  occurs,  the  indicated  operations  must  not 
be  performed  in  order,  from  left  to  right — the  multiplication 
must  be  performed  first.  Thus,  24x3-=-4x2-h9x5  =  f 
Ans.  In  order  to  obtain  the  same  result  that  would  be 
obtained  by  performing  the  indicated  operations  in  order, 
from  left  to  right,  symbols  of  aggregation  must  be  used. 
Thus,  by  using  two  vinculums,  the  last  expression  becomes 
24  x  3-4-4  X  2  -=-  9X  5  =  20,  the  same  result  that  would  be 
obtained  by  performing  the  indicated  operations  in  order, 
from  left  to  right.  


EXAMPLES   FOR    PRACTICE. 

1  74.      Find  the  values  of  the  following  expressions  : 


(a)  (8  +  5  -  1)  -5-  4. 

(b)  5  X  34  -  32. 

(c)  5x24  -=-15. 

(d)  144-5x24. 

(*)  (1,691-540  +  559)-!- 3X57. 

(/  )  2,080  +  120  -  80  x  4  -  1,670. 

(g)  (90  +  60  --  25)  X  5  -  29. 

(k)       90  +  60  -«-  25  X  5. 


(a)  3. 

(*)  88. 

(c)  8. 

(d)  24. 

(e)  10. 
(/)  210. 


1.2. 


ARITHMETIC. 

(CONTINUED.) 


PERCENTAGE. 

175.  Percentage    is   the    process   of   calculating    by 
hundredth*. 

1 76.  The  term  per  cent,  is  an  abbreviation  of  the  Latin 
words  per  centum,  which  mean  by  the  Jiundred.     A  certain 
per  cent,  of  a  number  is  the  number  of  hundreds  of  that 
number  which  is  indicated  by  the  number  of  units  in  the 
per  cent.      Thus,  6  per  cent,  of  125  is  125  X  yfj-  =  7.5  ;  25 
per  cent,  of  80  is  80  X  T8A  —  20  ;  43  per  cent,  of  432  pounds 
is  432  X  T4o3o  =  185.76  pounds. 

17^7.  The  sign  of  per  cent,  is  yd,  and  is  read  per  cent. 
Thus,  6$  is  read  six  per  cent.;  12£*  is  read  twelve  and  one-half 
per  cent. ,  etc. 

When  expressing  the  per  cent,  of  a  number  to  use  in  cal- 
culations it  is  customary  to  express  it  decimally  instead  of 
fractionally.  Thus,  instead  of  expressing  6$,  25$,  and  43$ 
as  yl^,  y2^,  and  y4^,  it  is  usual  to  express  them  as  .06,  .25, 
and  .43. 

The  following  table  will  show  how  any  per  cent,  can  be 
expressed  either  as  a  decimal  or  as  a  fraction : 

TABLE    3. 


Per  Cent.       Decimal.      Fraction.       Per  Cent.      Decimal.     Fraction. 


1*.... 
2*.... 

10*.... 

25*.... 

50*... . 

75*.... 
.100*.... 
125* .... 


.01 

.02 

.05 

.10 

.25 

.50 

.75 

1.00 

1.25 


T7Aor| 
{•U  or  1 


150  * 

500  * 

i*. 

12i*. 

624*. 


1.50 

5.00 
.0025 
.005 
.015 
.08^ 
.125 

.625 


T\>*0 


56  ARITHMETIC. 

178.  The  names  of  the  different  elements  used  in  per- 
centage are  :  the  base,  the  rate  per  cent.,  the  percentage,  the 
amount,  and  the  difference. 

1 79.  The  base  is  the  number  on  which  the  per  cent,  is 
computed. 

180.  The  rate  is  the  number  of  hundredths  of  the  base 
to  be  taken. 

181.  The  percentage  is  the  part,  or  number  of  hun- 
dredths, of  the  base  indicated  by  the  rate  ;  or  the  percent- 
age is  the  result  obtained  by  multiplying  the  base  by  the  rate. 

Thus,  when  it  is  stated  that  1%  of  $25  is  $1.75,  $25  is  the 
base,  7#  is  the  rate,  and  $1.75  is  the  percentage. 

1 82.  The  amount  is  the  sum  of  the  base  and  percentage. 

1 83.  The  difference  is  the  remainder  obtained  by  sub- 
tracting the  percentage  from  the  base. 

Thus,  if  a  man  has  $180,  and  he  earns  6#  more,  he  will 
have  altogether  $180  +  $180 X. 06,  or  $180 +  $10. 80  =  $190.80. 
Here  $180  is  the  base,  6#  the  rate,  $10.80  the  percentage, 
and  $190.80  the  amount. 

Again,  if  an  engine  of  125  horsepower  uses  16$  of  it  in 
overcoming  friction  and  other  resistances,  the  amount  left 
for  performing  useful  work  is  125  —  125  x  .16  =  125  —  20  = 
105  horsepower.  Here  125  is  the  base,  16#  the  rate,  20 
the  percentage,  and  105  the  difference. 

184.  From  the  foregoing  it  is  evident  that,  to  find  the 
percentage,  the  base  must  be  multiplied  by  the  rate.      Hence 
the  following 

Rule  21. —  To  find  the  percentage,  multiply  the  base  by 
the  rate,  expressed  decimally. 

EXAMPLE.— Out  of  a  lot  of  300  boiler  tubes  76#  were  used  in  a  boiler. 
How  many  tubes  were  used  ? 

SOLUTION.— 76^,  the  rate,  expressed  decimally,  is  .76  ;  the  base  is 
300 ;  hence,  the  number  of  tubes  used,  or  the  percentage,  is  by  the 
above  rule  300  X  •  76  =  228  tubes.  Ans. 

Expressing  the  rule  as  a  formula,  we  have 
percentage  =  base  X  rate. 


ARITHMETIC.  57 

185.  When  the  percentage  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  percentage  by  the  rate.     For, 
suppose  that  12  is  6$,  or  y^,  of  some  number;  then,  \%,  or 
yffr,  of  the  number  is  12  -^  6  or  2.      Consequently,  if  2  =1#, 
or  yffr,  100#,  or  $$$  =  2  X  100  =  200.     But,  since  the  same 
result  may  be  arrived  at  by  dividing  12  by  .06,  for  12  -f-  .06  = 
200  it  follows  that 

Rule  22. —  When  the  percentage  and  rate  are  given,  to 
find  the  base,  divide  the  percentage  by  the  rate,  expressed 
decimally. 

Formula,  base  =  percentage  -4-  rate. 

EXAMPLE.—  76£  of  a  lot  of  boiler  tubes  are  used  in  the  construction 
of  a  boiler.  If  the  number  of  tubes  used  were  228,  how  many  tubes 
were  in  the  lot  ? 

SOLUTION. — Here  228  is  the  percentage,  and  76£,  or  .76,  is  the  rate; 
hence,  applying  the  rule, 

228  -r- .  76  =  300  tubes.     Ans. 

186.  When  the  base  and  percentage  are  given  to  find 
the  rate,  the  rate  may  be  found,  expressed  decimally,  by 
dividing  the  percentage  by  the  base.      For  suppose  that  it 
is  desired  to  find  what  per  cent.  12  is  of  200.     \%  of  200  is 
200  X  .01  =  2.     Now,  if  1$  is  2,  12  is  evidently  as  many  per 
cent,  as  the  number  of  times  that  2  is  contained  in  12,  or  12  -=- 
2  =  6$.      But  the  same  result  may  be  obtained  by  dividing 
12,  the  percentage,  by  200,  the  base,  since  12  -^  200=. 06  =  6#. 
Hence, 

Rule  23. —  When  the  percentage  and  base  are  given,  to  find 
the  rate,  divide  the  percentage  by  the  base,  and  the  result 
ivill  be  the  rate,  expressed  decimally. 

Formula,  rate  =  percentage  -f-  base. 

EXAMPLE.— Out  of  a  lot  of  300  boiler  tubes,  228  were  used.  What 
per  cent,  of  the  total  number  was  used  ? 

SOLUTION.— Here  300  is  the  base  and  228  the  percentage ;  hence, 
applying  rule, 

rate  =  228  -r-  300  =  .  76  =  76£.     Ans. 

EXAMPLE.— What  per  cent,  of  875  is  25  ? 

SOLUTION.— Here  875  is  the  base,  and  25  is  the  percentage ;  hence, 
applying  rule, 

25 -5- 875  =.02?  =  2?^.     Ans. 

PROOF.— 875  X  02f  =  25. 


58  ARITHMETIC. 

EXAMPLES  FOR  PRACTICE. 

1 87.      What  per  cent,  of 


(a)  360  is  90? 
(&)  900  is  360? 

(c)  125  is  25? 

(d)  150  is  750? 

(e)  280  is  112  ? 
(/)  400  is  200  ? 
(g)  47  is  94  ? 


Ans. 


(a)  25*. 
(6)  40*. 

(c)  20*. 

(d)  500*. 
(*)  40*. 
(/)  50*. 
(g)  200*. 
(k)  50*. 


(A)  500  is  250?  

1 88.  The  amount  may  be  found  when  the  base  and  rate 
are  given,  by  multiplying  the  base  by  1  plus  the  rate,  ex- 
pressed decimally.      For  suppose  that  it  is  desired  to  find 
the  amount  when  200  is  the  base  and  6$  is  the  rate.     The 
percentage  is  200  X  .06  =  12,  and,  according  to  definition, 
Art.    182,  the  amount  is  200+12  —  212.     But  the  same 
result  may  be  obtained  by  multiplying  200  by  1  +  .06,  or 
1.06,  since  200  X  1.06  =  212.     Hence, 

Rule  24. —  When  the  base  and  rate  are  given,  to  find  the 
amount,  multiply  the  base  by  1  plus  the  rate,  expressed  deci- 
mally. 

Formula,  amount  —  base  X  (1  +  rate}. 

EXAMPLE.— If  a  man  earned  $725  in  a  year,  and  the  next  year  10* 
more,  how  much  did  he  earn  the  second  year  ? 

SOLUTION. — Here  725  is  the  base  and  10*  is  the  rate,  and  the  amount 
is  required.  Hence,  applying  the  rule, 

725  X  1  •  1 0  =  $797. 50.     Ans. 

189.  When  the  base  and  rate  are  given,  the  difference 
may  be  found  by  multiplying  the  base  by  1  minus  the  rate, 
expressed  decimally.     For  suppose  that  it  is  desired  to  find 
the  difference  when  the  base  is  200  and  the  rate  is  6#.     The 
percentage  is  200  X  .06  =  12  ;  and,  according  to  definition, 
Art.  183,  the  difference  =  200  —  12  =  188.     But  the  same 
result  may  be  obtained  by  multiplying  200  by  1  —.06,  or  .94, 
since  200  X  .94  =  188.     Hence, 

Rule  25. —  When  the  base  and  rate  are  given,  to  find  the 
difference,  multiply  the  base  by  1  minus  the  rate,  expressed 
decimally. 

Formula,  difference  =  base  X  (1  —  rate}. 


ARITHMETIC.  59 

EXAMPLE.— Out  of  a  lot  of  300  boiler  tubes  all  but  24#  were  used  in 
one  boiler ;  how  many  tubes  were  used  ? 

SOLUTION.— Here  300  is  the  base,  24£  is  the  rate,  and  it  is  desired  to 
find  the  difference.  Hence,  applying  the  rule. 

300  X  (1-. 24)  =  228  tubes.     Ans. 

190.  When  the  amount  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  amount  by  1  plus  the  rate. 
For  suppose  that  it  is  known  that  212  equals  some  number 
increased  by  6$  of  itself.      Then,  it  is  evident  that  212  equals 
106$  of  the  number  (base)  that  it  is  desired  to  find.     Con- 
sequently, if  212  =  106$,  1$  =  fif  =  2,  and  100$  =  2  X  100  = 
200  =  the  base.      But  the  same  result  may  be  obtained  by 
dividing    212    by    1  +  .06   or   1.06,  since    212-^1.06  =  200. 
Hence, 

Rule  26. —  When  the  amount  and  rate  arc  given,  to  find 
the  base,  divide  the  amount  by  1  plus  the  rate,  expressed 
decimally.  * 

Formula,  base  =  amount  -t-  (1  -{-  rate). 

EXAMPLE. — The  theoretical  discharge  of  a  certain  pump  when  run- 
ning at  a  piston  speed  of  100  feet  per  minute  is  278,910  gallons  per  day 
of  10  hours.  Owing  to  leakage  and  other  defects  this  value  is  25$ 
greater  than  the  actual  discharge.  What  is  the  actual  discharge  ? 

SOLUTION.— Here  278,910  equals  the  actual  discharge  (base)  increased 
by  25#  of  itself.  Consequently,  278,910  is  the  amount,  25£  is  the  rate, 
and  applying  rule, 

actual  discharge  =  278,910  -H  1.25  =  223,128  gallons.     Ans. 

191.  When  the  difference  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  difference  by  1  minus  the  rate. 
For  suppose  that  188  equals  some  number  less  6$  of  itself. 
Then,  188  evidently  equals  100  —  6  =  94$  of  some  number. 
Consequently,  if  188  =  94$,   1$  =  188  -=-  94  =  2,  and  100$  = 
2  X  100  =  200.     But   the    same   result  may  be  obtained    by 
dividing  188  by  1  —.06,  or  .94,  since  188  -4-  .94  =  200.    Hence, 

Rule  27. —  When  the  difference  and  rate  are  given,  to  find 
the  base,  divide  the  difference  by  1  minus  the  rate,  expressed 
decimally. 

Formula,  base  =  difference  -f-  (1  —  rate). 


60  ARITHMETIC. 

EXAMPLE.  —From  a  lot  of  boiler  tubes  76£  were  used  in  the  construc- 
tion of  a  boiler.  If  there  were  72  tubes  unused,  how  many  tubes  were 
in  the  lot  ? 

SOLUTION.— Here  72  is  the  difference  and  10%  is  the  rate.     Applying 

rule, 

73 .5-  (1  _  .  76)  =  300  tubes.     Ans. 

EXAMPLE. — The  theoretical  number  of  foot-pounds  of  work  per 
minute  required  to  operate  a  boiler  feed-pump  is  127,344.  If  30^  of  the 
total  number  actually  required  be  allowed  for  friction,  leakage,  etc., 
how  many  foot-pounds  are  actually  required  to  work  the  pump  ? 

SOLUTION.— Here  the  number  actually  required  is  the  base  ;  hence, 
127,344  is  the  difference,  and  30£  is  the  rate.  Applying  the  rule, 

127, 344  -4-  (1  -  .  30)  =  181 , 920  foot-pounds.     Ans. 

192.  EXAMPLE. — A  certain  chimney  gave  a  draft  of  2.76  inches 
of  water.     By  increasing  the  height  20  feet  the  draft  was  increased  to 
3  inches  of  water.     What  was  the  gain  per  cent.  ? 

SOLUTION.— Here  it  is  evident  that  3  inches  is  the  amount  and  that 
2.76  inches  is  the  base.  Consequently,  3 -2. 76  =  .24  inch  is  the 
percentage,  and  it  is  required  to  find  the  rate.  Hence,  applying 
rule  23, 

gain  per  cent.  =  .24  -=-  2.76  =  .087  =  8.1%.     Ans. 

193.  EXAMPLE. — A  certain  chimney  gave  a  draft  of  3  inches  of 
water.     After  an  economizer  had  been  put  in  the  draft  was  reduced  to 
1.2  inches  of  water.     What  was  the  loss  per  cent.? 

SOLUTION. — Here  it  is  evident  1.2  inches  is  the  difference  (since  it 
equals  3  inches  diminished  by  a  certain  per  cent,  loss  of  itself)  and  3 
inches  is  the  base.  Consequently,  3  —  1.2  =  1.8  inches  is  the  percent- 
age. Hence,  applying  rule  23, 

loss  per  cent.  =  1.8  -5-  3  =  .60  =  60£.     Ans. 

194.  To  find  the  gain  or  loss  per  cent.: 

Rule  28. — Find  the  difference  between  the  initial  and 
final  values  ;  divide  this  difference  by  the  initial  value. 

EXAMPLE. — If  a  man  buys  a  steam  engine  for  $1,860  and  some  time 
afterwards  purchases  a  condenser  for  25g  of  the  cost  of  the  engine, 
does  he  gain  or  lose,  and  how  much  per  cent.,  if  he  sells  both  engine 
and  condenser  for  §2,100  ? 

SOLUTION.— The  cost  of  the  condenser  was  §1,860  X  .25  =  $465;  con- 
sequently, the  total  initial  value,  or  cost,  was  $1,860  +  $465  =  $2,325. 
Since  he  sold  them  for  $2,100,  he  lost  $2,325  -  §2,100  =  §225.  Hence, 
applying  rule, 

225  -r-  2, 325  =  .  0968  =  9. 68£  loss.     Ans. 


ARITHMETIC.  61 


EXAMPLES   FOR   PRACTICE. 

195.      Solve  the  following: 


(a)  What  is  124$  of  §900  ? 
(6)  "  "  |$  "  627? 
(c)  "  "  33£$  "  54? 
(d)  101  is  68f$  of  what  number?  . 

(e)   784  "  83i$  ••      " 


(a)   $112.50. 
(t)    5.016. 
(O    18- 
(X) 


(«?)   940.8. 


(//)   40$. 


(/)  What  $  of  960  is  160? 
(£-)       "      "  "  $3, 606  is  $450|? 
(X)        "      "  "  280  is  112? 

1.  A  steam  plant  consumed  an  average  of  3,640  pounds  of  coal  per 
day.     The  engineer  made  certain  alterations  which  resulted  in  a  sav- 
ing of  250  pounds  per  day.     What  was  the  per  cent,  of  coal  saved  ? 

Ans.  7$,  nearly. 

2.  If  the  speed  of  an  engine  running  at  126  revolutions  per  minute 
should  be  increased  64$,  how  many  revolutions  per  minute  would  it 
then  make  ?  Ans.  134.19  revolutions. 

3.  The  list  price  of  an  engine  was  $1,400  ;  of  a  boiler,  $1,150,  and  of 
the  necessary  fittings  for  the  two,  $340.     If  25$  discount  was  allowed 
on  the  engine,  22$  on  the  boiler,  and  124$  on  the  fittings,  what  was  the 
actual  cost  of  the  plant  ?  Ans.  $2,244.50. 

4.  If  I  lend  a  man  $1,100,  and  this  is  184$  of  the  amount  that  I  have 
on  interest,  how  much  money  have  I  on  interest  ?  Ans.  $5,945.95. 

5.  A  test  showed  that  an  engine  developed  190.4  horsepower,  15$  of 
which  was  consumed  in  friction.     How  much    power  was   available 
for  use  ?  Ans.  161.84  H.  P. 

6.  By  adding  a  condenser   to  a  steam   engine,  the  power  was  in- 
creased 14$,and  the  consumption  of  coal  per  horsepower  per  hour  was 
decreased  20$.     If  the  engine  could  originally  develop  50  horsepower 
and  required  34  pounds  of  coal  per  horsepower  per  hour,  what  would 
be   the  total  weight  of  coal   used   in  an  hour,  with   the  condenser, 
assuming  the  engine  to  run  full  power  ?  Ans.  159.6  pounds. 


DENOMINATE  NUMBERS. 

196.  A  denominate  number  is  a  concrete  number, 
and  may  be  either  simple  or  compound,  as  8  quarts,  5  feet, 
ten  inches,  etc. 

197.  A    simple    denominate    number   consists   of 
units  of   but    one  denomination,  as  16  cents,    10  hours,  5 
dollars,  etc. 


62  ARITHMETIC. 

198.  A    compound    denominate  number  consists 
of  units  of  two  or  more  denominations  of  a  similar  kind,  as 
3  yards,  2  feet,  1  inch  ;  3  pounds,  5  ounces. 

199.  In  whole  numbers  and  in  decimals,  the  law 

of  increase  and  decrease  is  on  the  scale  of  10,  but  in  com- 
pound or  denominate  numbers  the  scale  varies. 


MEASURES. 

200.  A  measure  is  a  standard  unit  established  by  law 
or  custom,  by  which  quantity  of  any  kind  is  measured.     The 
standard  unit  of  dry  measure  is  the  Winchester  bushel ; 
of  weight,  the  pound ;  of  liquid  measure,  the  gallon,  etc. 

201.  Measures  are  of  six  kinds: 

1.  Extension.  4.     Time. 

2.  Weight.  5.     Angles. 

3.  Capacity.  6.     Money  or  value. 


MEASURES    OF    EXTENSION. 

2O2.     Measures  of  extension  are  used  in  measuring 
lengths,  distances,  surfaces,  and  solids. 


LINEAR   MEASURE. 

TABLE   4. 


12    inches  (in.)  =  1  foot    . 

3    feet  =  1  yard    . 

5£  yards  =  1  rod      . 

40  rods  =  1  furlong 

8  furlongs        =  1  mile    . 


Abbreviation, 
ft. 
yd. 
rd. 
fur. 
mi. 


yd. 


rd.       fur. 


in.  ft. 

36=        3 

198  =      16i=        5| 
7,920  =    660  =    220  =  40 

=5,280  =1,760  =320=    8 


SQUARE   MEASURE. 

TABLE   5. 

144  square  inches  (sq.  in.)  . 
9  square  feet 
30£  square  yards 
160  square  rods 
640  acres. 


sq.  mi. 


A. 


5  (sq.  in.) 

.     .    = 

1  square  foot 
1  square  yard 
1  square  rod 

sq.  rd. 

sq.  yd. 

1  acre   .     .    . 
1  square  mile 
sq.  ft. 

sq.  ft. 
sq.  yd. 
sq.  rd. 

.  A. 
sq.  mi. 


sq.  in. 


1       =  640  =  102,400  =  3,097,600  =  27,878,400  =  4,014,489,600 


ARITHMETIC.  63 

CUBIC  MEASURE. 

TABLE   6. 

1728    cubic  inches  (cu.  in.)  .     .     .    =    1  cubic  foot cu.  ft. 

27    cubic  feet =1  cubic  yard cu.  yd. 

128    cubic  feet =    1  cord cd. 

24f  cubic  feet =1  perch P. 

cu.  yd.  cu.  ft.         cu.  in. 
1     =     27     =     46,656 


MEASURES    OF    WEIGHT. 


AVOIRDUPOIS  WEIGHT. 

TABLE   7. 

16  ounces  (oz.) =    1  pound Ib. 

100  pounds =    1  hundred-weight.     .     .     .  cwt. 

20  cwt.,  or  2,000  Ib =    1  ton T. 

T.        cwt.  Ib.                 oz. 

1     =     20     =  2,000     =     32,000 

203.  The   ounce  is  divided   into  halves,  quarters,  etc. 
Avoirdupois  weight  is  used  for  weighing  coarse  and  heavy 
articles. 

LONG  TON   TABLE. 

TABLE  8. 

16  ounces =1  pound Ib. 

112  pounds =    1  hundred-weight.     .     .     .  cwt. 

20  cwt,  or  2,240  Ib =1  ton T. 

In  all  the  calculations  throughout  this  and  the  succeeding 
volumes,  2,000  pounds  will  be  considered  one  ton,  unless  the 
long  ton  (2,240  pounds)  is  especially  mentioned. 

TROY  WEIGHT. 

TABLE   9. 

24  grains  (gr.) =    1  pennyweight    ....     pwt. 

20  pennyweight =1  ounce oz. 

12  ounces =1  pound Ib. 

Ib.          oz.  pwt.             gr. 

1     =     12     =  240     =     5,760 

204.  Troy  weight  is  used  in  weighing  gold  and  silver 
ware,  jewels,  etc.     It  is  used  by  jewelers. 


64 


ARITHMETIC. 
MEASURES    OF    CAPACITY. 


LIQUID  MEASURE. 

TABLE    10. 

Pt 

gins  (gi.)  
2    pints    
4    quarts  
31|  gallons      
63    gallons      
hhd.      bbl.        gal. 
1     =     2     =     63 

—    1  quart 

.     .     qt. 
.     .   gal. 
.     .  bbl. 
.  hhd. 

.    =    1  gallon     
.    =    1  barrel     

qt.             pt.               gi. 
=     252     =     504     =     2,016 

2  pints  (pt.) 
8  quarts  . 
4  pecks.  . 


DRY   MEASURE. 

TABLE    11. 

=    1  quart . 

=    1  peck  . 

=1  bushel 

bu.         pk.         qt.  pt. 

1     =     4    =     32     =     64 


qt. 
pk. 

bu. 


60  seconds  (sec.) 
60  minutes 

MEASURE    OF    TIME. 

TABLE   12. 

=r     1  minute  

24  hours   .     .     . 

—     1  day 

7  days     .     .     . 
365  days       j 
12  months  j 
366  days     .     .     . 
100  years   .     .     . 

.     =     1  week      ...... 
=     1  common  year     .     .     . 

.   "  =     1  leap  year. 
=     1  century. 

mm. 
hr. 


wk. 


NOTE. — It  is  customary  to  consider  one  month  as  30  days. 


MEASURE  OF  ANGLES  OR  ARCS. 

TABLE   13. 

60  seconds  (") =  1  minute ' 

60  minutes =  1  degree *  . 

90  degrees =  1  right  angle  or  quadrant  [_  • 

360  degrees =  1  circle cir. 

cir. 

1     =     360°     =     21,600'  =     1,296,000" 


ARITHMETIC.  65 


MEASURE    OF    MONEY. 

UNITED  STATES  MONEY. 

TABLE    14. 

.   =     1  cent ct. 

1  dime d. 

1  dollar $. 

1  eagle E. 

ct.  m. 

1,000      ::      10,000 

MISCELLANEOUS    TABLE. 

TABLE    15. 


10  mills  (m.)  .     . 
10  cents 

.     .     . 

.     .     . 

•   = 

10  dimes    .     .     . 
10  dollars  .     .     . 

.   = 

E. 
1 

$ 
=     10 

d. 
=     100 

-    1 

12  things  are  1  dozen. 

12  dozen  are  1  gross. 

12  gross  are  1  great  gross. 

2  things  are  1  pair. 
20  things  are  1  score. 

1  league  is  3  miles. 

1  fathom  is  6  feet. 


1  meter  is  nearly  39.87  inches. 

1  hand  is  4  inches. 

1  palm  is  3  inches. 

1  span  is  9  inches. 
24  sheets  are  1  quire. 
20  quires,  or  480  sheets,  are  1  ream. 

1  bushel  contains  2,150.4  cubic  in. 


1  U.  S.  standard  gallon  (also called  a  wine  gallon)  contains  231  cubic  in 
1  U.  S.  standard  gallon  of  water  weighs  8.355  pounds,  nearly. 
1  cubic  foot  of  water  contains  7.481  U.  S.  standard  gallons,  nearly. 
1  British  imperial  gallon  weighs  10  pounds. 

It  will  be  of  great  advantage  to  the  student  to  carefully 
memorize  all  of  the  above  tables. 


REDUCTION    OF    DENOMINATE    NUMBERS. 

205.  Reduction  of  denominate  numbers  is  the    pro- 
cess of  changing  their  denomination  without  changing  their 
value.     They  may  be  changed  from  a  higher  to  a  lower 
denomination  or  from  a  lower  to  a  higher — either  is  reduc- 
tion.    As, 

2  hours  =  120  minutes. 
32  ounces  =  2  pounds. 

206.  Principle. — Denominate   numbers   are   changed 
to  lower  denominations  by  multiplying,   and  to  Jiiglier  de- 
nominations by  dividing. 

To  reduce  denominate  numbers  to  louver  denomi- 
nations : 


66  ARITHMETIC. 

2O7.     EXAMPLE.— Reduce  5  yd.  2  ft.  7  in.  to  inches. 
SOLUTION.—  yd.          ft.  in. 


15ft. 

2ft. 

17ft. 

12 

"34 
17 

2  0  4  in. 
'7  in. 

211  inches.     Ans. 

EXPLANATION. — Since  there  are  3  feet  in  1  yard,  in  5  yards 
there  are  5  X  3,  or  15  feet,  and  15  feet  plus  2  feet  -  17  feet. 
There  are  12  inches  in  a  foot  ;  therefore,  12  X  17  =  204 
inches,  and  204  inches  plus  7  inches  =  211  inches  =  number 
of  inches  in  5  yards  2  feet  and  7  inches.  Ans. 

2O8.      EXAMPLE. — Reduce  6  hours  to  seconds. 
SOLUTION. —  6        hours. 

60 

360      minutes. 
60 


21600  seconds.     Ans. 

EXPLANATION. — As  there  are  60  minutes  in  one  hour,  in 
six  hours  there  are  6  X  60,  or  360  minutes  ;  as  there  are  no 
minutes  to  add,  we  multiply  360  minutes  by  60,  to  get  the 
number  of  seconds. 

209.  In  order  to  avoid  mistakes,  if  any  denomination 
be  omitted,  represent  it  by  a  cipher.      Thus,  before  reducing 
3  rods  6  inches  to  inches,  insert  a  cipher  for  yards  and  a 
cipher  for  feet ;  as, 

rd.        yd.        ft.        in. 
3006 

21 0.  Rule. — Multiply  the  number  representing  the  high- 
est denomination  by  the  number  of  units  in  the  next  lower 


ARITHMETIC.  67 

required  to  make  one  of  the  higher  denomination,  and  to  the 
product  add  the  number  of  given  units  of  that  lower  denomi- 
nation. Proceed  in  this  manner  until  the  number  is  reduced 
to  the  required  denomination. 


EXAMPLES  FOR   PRACTICE. 

211.     Reduce 


(a)  4  rd.  2  yd.  2  ft.  to  ft. 

(£)  4  bu.  3  pk.  2  qt.  to  qt. 

(<:)  13  rd.  5  yd.  2  ft.  to  ft. 

(d)  5  mi.  100  rd.  10  ft.  to  ft. 

(e)  8  Ib.  4  oz.  6  pwt.  to  gr. 
(/)  52  hhd.  24  gal.  1  pt,  to  pt. 
(g )  5  cir.  16°  20'  to  minutes. 
(K)  14  bu.  to  qt. 


Ans. 


(a)  74ft 

(b)  154  qt. 

(c)  231.5  ft. 

(d)  .28,060ft. 

(e)  48,144gr. 
(/)  26,401  pt. 
(g)  108,980'. 
(K)  448  qt. 


To  reduce  lower  to  higher  denominations: 

212.  EXAMPLE. — Reduce  211  in.  to  higher  denominations. 
SOLUTION.—  1  2  )  2  1 1  in. 

3  )  1  7  ft.  +  7  in. 

5  yd.  -(-  2  ft.     Ans. 

EXPLANATION. — There  are  12  inches  in  1  foot  ;  there- 
fore, 211  divided  by  12  =  17  feet  and  7  inches  over.  There 
are  3  ft.  in  1  yd. ;  therefore,  17  ft.  divided  by  3  =  5  yd.  and 
2  ft.  over.  The  last  quotient  and  the  two  remainders  con- 
stitute the  answer,  5  yd.  2  ft.  7  in. 

213.  EXAMPLE. — Reduce  14,135  gi.  to  higher  denominations. 
SOLUTION.—  4  )14135 

2  )    3533  pt.  3  gi. 
4 )    1766  qt.  1  pt. 
441  gal.  2qt. 
31.5)441.0(14bbl. 
315 

T¥eo 

1260 


EXPLANATION. — There  are  4  gi.  in  1  pt.,  and  in  14,135  gi. 
there  are  as  many  pints  as  4  is  contained  in  14,135,  or  3,533 


68  ARITHMETIC. 

pt.  and  3  gi.  remaining.  There  are  2  pt.  in  1  qt.,  and  in 
3,533  pt.  there  are  1,766  qt.  and  1  pt.  remaining.  There 
are  4  qt.  in  1  gal.,  and  in  1,766  qt.  there  are  441  gal.  and 
2  qt.  remaining.  There  are  31 J  gal.  in  1  bbl.,  and  in 
441  gal.  there  are  14  bbl. 

The  last  quotient  and  the  three  remainders  constitute  the 
answer,  14  bbl.  2  qt.  1  pt.  3  gi. 

214.  Rule  3O. — Divide  the  number  representing  tJie  de- 
nomination given,  by  Hie  ?tumber  of  units  of  tJiis  denomi- 
nation required  to  make  one  unit  of  the  next  higher  denomi- 
nation. The  remainder  will  be  of  the  same  denomination, 
but  the  quotient  will  be  of  the  next  higlier.  Divide  this 
quotient  by  the  number  of  units  of  its  denomination  required 
to  make  one  unit  of  the  next  higher.  Continue  until  the 
highest  denomination  is  reached,  or  until  there  is  not  enough 
of  a  denomination  left  to  make  one  of  the  next  higher.  The 
last  quotient  and  the  remainders  constitute  the  required  result. 


EXAMPLES  FOR   PRACTICE. 

21 5.     Reduce  to  units  of  higher  denominations: 
(a)  7,460  sq.  in. ;    (b)  7,580  sq.  yd. ;    (<r)   148,760  cu.  in. ;    (d)  17,651"  ; 
(<?)  8,000  gi. ;   (/)  36,450  Ib. 

(a)  5  sq.  yd.  6  sq.  ft.  116  sq.  in. 

(£)  1  A.  90  sq.  rd.  17  sq.  yd.  4  sq.  ft.  72  sq.  in. 


Ans. 


(<r)  3  cu.  yd.  5  cu.  ft.  152  cu.  in. 
(</)  4°  54'  11". 
(e)  3  hhd.  61  gal. 
(/)  18  T.  4  cwt.  50  Ib. 


ADDITION    OF    DENOMINATE    NUMBERS. 

216.      EXAMPLE. — Find  the  sum    of    3  cwt.  46  Ib.  12  oz. ;  8  cwt. 
12  Ib.  13  oz.  ;  12  cwt.  50  Ib.  13  oz. ;  27  Ib.  4  oz. 
SOLUTION.— 


T. 

cwt. 

Ib. 

oz. 

0 

3 

46 

12 

0 

8 

12 

13 

0 

12 

50 

"13 

0 

0 

27 

4 

37  10    Ans. 


ARITHMETIC.  69 

EXPLANATION. — Begin  to  add  at  the  right-hand  column  : 
4  +  13  +  13  +  12  =  42  ounces  ;  as  16  ounces  make  1  pound, 
42  ounces  -=-16  =  2  and  a  remainder  of  10  ounces,  or  2 
pounds  and  10  ounces.  Place  10  ounces  under  ounce 
column  and  add  2  pounds  to  the  next  or  pound  column. 
Then,  2  +  27+50  +  12  +  46=137  pounds;  as  100  pounds 
make  a  hundred-weight,  137  -r-  100  =  1  hundred-weight  and 
a  remainder  of  37  pounds.  Place  the  37  under  the  pound 
column,  and  add  1  hundred-weight  to  the  next  or  hundred- 
weight column.  Next,  1  +  12 +8  +  3  =  24  hundred-weight. 
20  hundred-weight  make  a  ton  ;  therefore  24  -f-  20  =  1  ton 
and  4  hundred-weight  remaining.  Hence,  the  sum  is  1  ton 
4  hundred -weight  37  pounds  10  ounces.  Ans. 

217.      EXAMPLE.— What  is  the  sum  of  2  rd.  3  yd.  2  ft.  5  in. ;  6  rd. 

1  ft.  10  in. ;  17  rd.  11  in. ;  4  yd.  1  ft.  ? 

SOLUTION.— 


rd. 

yd. 

ft. 

in. 

2 

3 

2 

5 

6 

0 

1 

10 

17 

0 

0 

11 

0 

4 

1 

0 

26 

3| 

0 

2 

or    26 

3 

1 

8    Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  the  first  col- 
umn =  26  inches,  or  2  feet  and  2  inches  remaining.  The 
sum  of  the  numbers  in  the  next  column  plus  2  feet  =  6  feet, 
or  2  yards  and  0  feet  remaining.  The  sum  of  the  next  col- 
umn plus  2  yards  =  9  yards,  or  9  -f-  5i  =  1  rod  and  3|  yards 
remaining.  The  sum  of  the  next  column  plus  1  rod  =  26 
rods.  To  avoid  fractions  in  the  sum,  the  £  yard  is  reduced 
to  1  foot  and  6  inches,  which  added  to  26  rods  3  yards  0  feet 
and  2  inches  =  26  rods  3  yards  1  foot  8  inches.  Ans. 

218.  EXAMPLE.— What  is  the  sum  of  47  ft.  and  3  rd.  2  yd.  2  ft. 
10  in.  ? 


70  ARITHMETIC. 

SOLUTION.— When  47  ft.  is  reduced  it  equals  2  rd.  4  yd.  2  ft.,  which 
can  be  added  to  3  rd.  2  yd.  2  ft.  10  in.     Thus, 

rd.  yd.  ft.  in. 

3  2  2  10 

2420 


6  1|  1  10 

or    6  2  0  4    Ans. 

219.  Rule  31. — Place  the  numbers  so  that  like  denomina- 
tions are  under  each  other.  Begin  at  the  right-hand  column, 
and  add.  Divide  the  sum  by  the  number  of  units  of  this 
denomination  required  to  make  one  unit  of  the  next  higher. 
Place  tlie  remainder  under  the  column  added,  and  carry  the 
quotient  to  the  next  column.  Continue  in  this  manner  until 
the  JiigJicst  denomination  given  is  reached. 


EXAMPLES   FOR   PRACTICE. 

22O.      What  is  the  sum  of 

(a)  25  Ib.  7  oz.  15  pwt  23  gr. ;  17  Ib.  16  pwt. ;  15  Ib.  4  oz.  12  pwt ; 
18  Ib.  16  gr. ;  10  Ib.  2  oz.  11  pwt.  16  gr.  ? 

(f>)  9  mi.  13  rd.  4  yd.  2  ft. ;  16  rd.  5  yd.  1  ft.  5  in. ;  16  mi.  2  rd.  3  in. , 
14  rd.  1  yd.  9  in.  ? 

(0    3  cwt.  46  Ib.  12  oz. ;  12  cwt.  9i  Ib. ;  2±  cwt.  21$  Ib  ? 

(d)  10  yr.  8  mo.  5  wk.  3  da. ;    42  yr.  6  mo.  7  da.  j    7  yr.  5  mo.  18  wk. 
4  da.;  17  yr.  17  da.? 

(e)  17  tons  11  cwt  49  Ib.  14  oz. ;  16  tons  47  Ib.  13  oz. ;  20  tons  13  cwt. 
14  Ib.  6  oz. ;  11  tons  4  cwt.  16  Ib.  12  oz.  ? 

(/)  14  sq.  yd.  8  sq.  ft.  19  sq.  in. ;  105  sq.  yd.  16  sq.  ft.  240  sq.  in.', 
42  sq.  yd.  28  sq.  ft.  165  sq.  in.  ? 

(a)   86  Ib.  3  oz   16  pwt  7  gr. 
(£)    25  mi.  47  rd.  1  ft.  5  in. 
(0    18  cwt  2  Ib.  14  oz, 
Ans'       (d)  78  yr.  1  mo.  3  wk.  3  da. 

(e)    65  tons  9  cwt.  28  Ib.  13  oz. 
(/)  167  sq.  yd.  136  sq.  in. 


SUBTRACTION  OF  DENOMINATE  NUMBERS. 

221.      EXAMPLE.— From  21  rd.  2  yd.  2  ft.  64  in.,  take  9  rd  4  yd. 
iin. 

SOLUTION.—  rd.         yd.         ft.          in. 

21  2  2  6i 

9  4  0          I0j 

11  34 I  8i        Ans. 


ARITHMETIC.  71 

EXPLANATION.  —  Since  10£  inches  can  not  be  taken  from 
6£  inches,  we  must  borrow  1  foot,  or  12  inches,  from  the  2  feet 
in  the  next  column  and  add  it  to  the  6f  G£  -f  12  =  18£. 
18£  inches  —  10£  inches  =  8J  inches.  Then,  0  foot  from  the 

1  remaining  foot  =  1  foot.     4  yards  can  not  be  taken  from 

2  yards;  therefore,  we  borrow  1  rod,  or  5^  yards,  from  21  rods 
and  add  it  to  2.     2  +  5£  —  7|  ;  7£  -  4  =  3^  yards.     9  rods 
from  20  rods  =  11  rods.      Hence,  the  remainder  is  11   rods 
3|  yards  1  foot  8£  inches.     Ans. 

To  avoid  fractions  as  much  as  possible,  we  reduce  the 
£  yard  to  inches,  obtaining  18  inches;  this  added  to  8£  inches, 
gives  26|  inches,  which  equals  2  feet  2{  inches.  Then, 
2  feet  +  1  foot  =  3  feet  =  1  yard,  and  3  yards  -f-  1  yard  =  4 
yards.  Hence,  the  above  answer  becomes  11  rods  4  yards 
0  feet  2{  inches. 

222.      EXAMPLE.  —What  is  the  difference  between  3  rd.  2  yd.  2  ft. 
10  in.  and  47  ft.  ? 
SOLUTION.—  47  ft.  =  2  rd.  4  yd.  2  ft. 

rd.         yd.         ft.         in. 
3  2  2         10 

2420 


0  3£          0          10 

or  324     Ans. 

To  find  (approximately)  the  interval  of  time  be- 
tween two  dates: 

223.  EXAMPLE.  —  How  many  years,  months,  days,  and  hours 
between  4  o'clock  P.M.  of  June  15,  1868,  and  10  o'clock  A.M.,  September  28, 
1891? 

SOLUTION.—  yr.        mo.        da.        hr. 

1891          8  28          10 

1868          5  15          16 


23          3  12          18    Ans. 

EXPLANATION.— Counting  24  hours  in  1  day,  4  o'clock 
P.M.  is  the  16th  hour  from  the  beginning  of  the  day,  or 
midnight.  On  September  28,  8  months  and  28  days  have 
elapsed,  and  on  June  15,  5  months  and  15  days.  After  plac- 
ing the  earlier  date  under  the  later  date,  subtract  as  in  the 
previous  problems.  Count  30  days  as  1  month. 


72  ARITHMETIC. 

224.  Rule  32. — Place  the  smaller  quantity  under  the 
larger  quantity,  with  like  denominations  under  eacli  other. 
Beginning  at  the  right,  subtract  successively  the  number  in  the 
subtrahend  in  each  denomination  from  the  one  above,  and  place 
the  differences  underneath.  If  the  number  in  the  minuend  of 
any  denomination  is  less  than  the  number  under  it  in  the  sub- 
trahend, one  must  be  borrowed  from  the  minuend  of  the  next 
higher  denomination,  reduced}and  added  to  it. 


EXAMPLES   FOR   PRACTICE. 
225.      From 

(a)  125  Ib.  8  oz.  14  pwt.  18  gr.  take  96  Ib.  9  oz.  10  pwt.  4  gr. 
(£)  126  hhd.  27  gal.  take  104  hhd.  14  gal.  1  qt.  1  pt. 

(c)  65  T.  14  cwt.  64  Ib.  10  oz.  take  16  T.  11  cwt.  14  oz. 

(d)  148  sq.  yd.  16  sq.  ft.  142  sq.  in.  take  132  sq.  yd.  136  sq.  In. 
(<?)  100  bu.  take  28  bu.  2  pk.  5  qt.  1  pt. 

(/)  14  mi.  34  rd.  16  yd.  13  ft.  11  in.  take  3  mi.  27  rd.  11  yd.  4  ft.  10  in. 

(a)  28  Ib.  11  oz.  4  pwt.  14  gr. 

(b)  22  hhd  12  gal.  2  qt.  1  pt. 
.    i  (c)  49  T.  3  cwt.  63  Ib.  12  oz. 

*  (d)  16  sq.  yd.  16  sq.  ft.  6  sq.  in. 
(e)  71  bu.  1  pk.  2  qt.  1  pt. 
(/)  11  mi.  7  rd.  5  yd.  9  ft.  1  in. 


MULTIPLICATION  OF  DENOMINATE  NUMBERS. 
226.     EXAMPLE.— Multiply  7  Ib.  5  oz.  13  pwt.  15  gr.  by  12. 
SOLUTION. —  Ib.        oz.        pwt.        gr. 

7  5  13  15 

12. 


89  8  3  12    Ans. 

EXPLANATION. — 15  grains  X  12  —  180  grains.  180  -=-  24  =  7 
pennyweights  and  12  grains  remaining.  Place  the  12  in  the 
grain  column  and  carry  the  7  pennyweights  to  the  next. 
Now,  13  X  12  +  7  =  103  pennyweights  ;  163  -4-  20  =  8  ounces 
and  3  pennyweights  remaining.  Then,  5  X  12  +  8  =  68 
ounces;  68  H-  12  —  5  pounds  and  8  ounces  remaining.  Then, 

7  X  12  +  5  =  89  pounds.      The  entire    product  is  89  pounds 

8  ounces  3  pennyweights  12  grains.     Ans. 


ARITHMETIC.  73 

227.  Rule  33. — Multiply  the  number  representing  each 
denomination  by  the  multiplier \  and  reduce  each  product  to  the 
next  higher  denomination,  writing  the  remainders  under  each 
denomination,  and  carrying  the  quotient  to  the  next,  as  in 
addition  of  denominate  numbers. 

22S.  NOTE.  —  In  multiplication  and  division  of  denominate 
numbers,  it  is  sometimes  easier  to  reduce  the  number  to  the  lowest 
denomination  given  before  multiplying  or  dividing,  especially  if  the 
multiplier  or  divisor  is  a  decimal.  Thus,  in  the  above  example,  had 
the  multiplier  been  1.2.  the  easiest  way  to  multiply  would  have  been  to 
reduce  the  number  to  grains,  then,  multiply  by  1.2,  and  reduce  the 
product  to  higher  denominations.  For  example,  7  Ib.  5oz.  13pwt.  15  gr. 
=  43,047  gr.  43,047  X  1.2  =  51,656.4  gr.  =  8  Ib.  11  oz.  12  pwt.  8.4  gr. 
Also.  43,047  X  12  =  516,564  gr.  =  89  Ib.  8  oz.  3  pwt.  12  gr.,  as  above. 
The  student  may  use  either  method. 


EXAMPLES   FOR   PRACTICE. 

229.  Multiply 

(«)  15  cwt.  90  Ib.  by  5;  (6)  12  yr.  10  mo.  4  wk.  3  da.  by  14;  (c)  11  mi. 
145  rd.  by  20;  (</)  12  gal.  4  pt.  by  9;  (e)  8  cd.  76  cu.  ft.  by  15;  (/)  4hhd. 
3  gal.  1  qt.  1  pt.  by  12. 

(a)  79  cwt.  50  Ib. 

(b)  180  yr.  11  mo.  2  wk. 
.        ,  (c)     229  mi.  20  rd. 

i1  (d)    112  gal.  2qt. 

(e)     128  cd.  116  cu.  ft. 
.(/)    48  hhd.  40  gal.  2  qt. 

DIVISION    OF    DENOMINATE    NUMBERS. 

230.  EXAMPLE.— Divide  48  Ib.  11  oz.  6  pwt.  by  8. 
SOLUTION.—  Ib.  oz.  pwt.         gr. 

8)48  11  6  0 

6  Ib.         1  oz.        8  pwt.      6  gr.     Ans. 

EXPLANATION.— After  placing  the  quantities  as  above, 
proceed  as  follows  :  8  is  contained  in  48  six  times  without  a 
remainder.  8  is  contained  in  11  ounces  once  with  3  ounces 
remaining.  3  X  20  =  60 ;  60  +  6  =  66.  pennyweights ;  66  pen- 
nyweights -7-8  =  8  pennyweights  and  2  remaining ;  2  X  24 
grains  =  48  grains;  48  grains  -*•  8  =  6  grains.  .  Therefore, 
the  entire  quotient  is  6  pounds  1  ounce  8  pennyweights  6 
grains.  Ans. 

EXAMPLE.— A  silversmith  melted  up  2  Ib.  8  oz.  10  pwt.  of  silver, 
which  he  made  into  6  spoons;  what  was  the  weight  of  each  spoon  ? 


74  ARITHMETIC. 

SOLUTION. —  Ib.          oz.  pwt. 

6)2  8  10 

5  oz.  8  pwt.     8  gr.     Ans. 

EXPLANATION. — Since  we  can  not  divide  2  pounds  by  6,  we 
reduce  it  to  ounces.  2  pounds  —  24  ounces,  and  24  ounces 
+  8  ounces  =  32  ounces ;  32  ounces  -4-  6  =  5  ounces  and 
2  ounces  over.  2  ounces  =  40  pennyweights.  40  penny- 
weights +  10  penny  weights  =  50  pennyweights,  and  50  pen- 
nyweights -4-6  =  8  pennyweights  and  2  pennyweights  over. 
2  pennyweights  =  48  grains,  and  48  grains  -4-  6  =  8  grains. 
Hence,  each  spoon  contains  5  ounces  8  pennyweights  8 
grains.  Ans. 


ARITHMETIC.  75 

231 .      EXAMPLE.— Divide  820  rd.  4  yd.  2  ft.  by  112. 

rd.      yd.  ft.    rd.  yd.  ft.     in. 
SOLUTION.—  112)820     4    2(7      1     2    5.143    Ans. 

784 

3  6  rd.  rem. 
5.5 

I¥o 

180 


198.0yd. 
4 

112)  2~02  yd.  ( 1  yd. 
112 

9  0  yd.  rem. 
3 

270  ft. 

2ft. 

112)  2^2  ft.  (  2  ft. 
224 

4  8  ft.  rem. 
12 

96 
48 

112)  5  76  in.  (5.1  42 8  + in.,  or  5.1  4 3  in. 
560 

112 

448 

320 
224 

~960 
896 
~64 

EXPLANATION. — The  first  quotient  is  7  rods  with  36  rods 
remaining.  5.5x36  =  198  yards;  198  yards -f  4  yards  = 
202  yards;  202  yards  -^  112  =  1  yard  and  90  yards  remain- 
ing. 90  X  3  =  270  feet ;  270  feet  +  2  feet  =  272  feet ;  272 
feet -T- 112  =  2  feet  and  48  feet  remaining;  48x12  =  576 
inches;  576  inches  -T-  112  =  5.143  inches,  nearly.  Ans. 


76  ARITHMETIC. 

The  preceding  example  is  solved  by  long  division,  because 
the  numbers  are  too  large  to  deal  with  mentally.  Instead 
of  expressing  the  last  result  as  a  decimal,  it  might  have 
been  expressed  as  a  common  fraction.  Thus,  576  -=-  112  — 
Sy1^  =  5|  inches.  The  chief  advantage  of  using  a  common 
fraction  is  that  if  the  quotient  be  multiplied  by  the  divisor, 
the  result  will  always  be  the  same  as  the  original  dividend. 

232.  Rule  34. — Find  how  many  times  the  divisor  is  con- 
tained in  the  first  or  highest  denomination  of  the  dividend. 
Reduce  tJie  remainder  (if  any)  to  the  next  lower  denomination, 
and  add  to  it  the  number  in  the  given  dividend  expressing  tJiat 
denomination.     Divide  this  new  dividend  by  the  divisor.      The 
quotient  will  be  the  next  denomination  in  the  quotient  required. 
Continue  in  this   manner   until  the  lowest  denomination   is 
reached.      The  successive  quotients  will  constitute  the  entire 
quotient. 

EXAMPLES   FOR   PRACTICE. 

233.  Divide 

(a)  376  mi.  276  rd.  by  22;  (b)  1,137  bu.  3  pk.  4  qt.  1  pt.  by  10;  (c)  84 
cwt.  48  Ib.  49  oz.  by  16;  (d)  78  sq.  yd.  18  sq.  ft.  41  sq.  in.  by  18;  (e)  148 
mi.  64  rd.  24  yd.  by  12;  (/)  100  tons  16  cwt.  18  Ib.  11  oz.  by  15;  (g)  36 
Ib.  18  oz.  18  pwt.  14  gr.  by  8;  (A)  112  mi.  48  rd.  by  100. 

(a)     17  mi.  41T7T  rd. 

(£)     113  bu.  3  pk.  1  qt.  }  pt. 

(c)  5  cwt.  28  Ib.  3TV  oz. 

(d)  4  sq.yd.  4  sq.ft.  2ysg  sq.in. 

(e)  12  mi.  112  rd.  2  yd. 

(/)     6  tons  14  cwt.  41  Ib.  3|£  oz. 
(£•)    4  Ib.  8  oz.  7  pwt.  7f  gr. 
(h)    1  mi.  38}f  rd. 


INVOLUTION. 

23-1.  Involution  is  the  process  of  multiplying  a  num- 
ber by  itself  one  or  more  times.  The  product  obtained  by 
multiplying  a  number  by  itself  is  called  a  power  of  that 
number. 

Thus,  the  second  power  of  3  is  9,  since  3x3  are  9. 


Ans. 


ARITHMETIC.  77 

The  third  power  of  3  is  27,  since  3  X  3  X  3  are  27. 

The  fifth  power  of  2  is  32,  since  2  X  2  X  2  x  2  X  2  are  32. 

235.  An  exponent  is  a  small  figu replaced  to  the  right 
and  a  little  above  a  number  to  show  to  what  flower  it  is  to  be 
raised,  or  how  many  times  the  number  is  to  be  used  as  a 
factor,  as  the  small  figures  2>  3>  and  s  below : 

3'  =  3  X  3  =  9. 

3°  =  3  X  3  X  3  =  27. 

2*  =  2  X  2  X  2  X  2  X  2  =  32. 

236.  The  root  of  a  number  is  that  number  which,  used 
the  required  number  of  times  as  a  factor,  produces  the  num- 
ber.    In  the  above  cases,  3  is  a  root  of  9,  since  3  X  3  are  9. 
It  is  also  a  root  of  27,  since  3x3x3  are  27.     Also,  2  is  a 
root  of  32,  since  2  X  2  x  2  X  2  X  2  are  32. 

237.  The  second  power    of    a    number    is    called    its 
square. 

Thus,  53  is  called  the  square  of  5,  or  5  squared,  and  its 
value  is  5  X  5  =  25. 

238.  The  third  power  of  a  number  is  called  its  cube. 
Thus,  53  is  called  the  cube  of  5,  or  5  cubed^  and  its  value  is 

5X5X5  =  125. 

To    find    any  power    of  a    number  : 

239.  EXAMPLE.— What  is  the  third  power,  or  cube,  of  36  ? 
SOLUTION.—  35x35x35, 

or      35 
35 

175 
105 


1225 
35 

6125 
3675 
=  42875  Ans. 


78  ARITHMETIC. 

24O.      EXAMPLE.  -What  is  the  fourth  power  of  15? 
SOLUTION.—  15x15x15x15, 

or  15 

15 

75 
15 


15 

1125 
225 

3375 
15 


16875 
3375 

fourth  power  =  50625    Ans. 

241 .  EXAMPLE.—    1.2s  =  what  ? 
SOLUTION.—  1.2x1.2x1.2, 

or       1.2 
1.2 

1~44 
1.2 

"288 
144 

1.728    Ans. 

242.  EXAMPLE.— What  is  the  third  power,  or  cube,  of  f  ? 
SOLUTION.-        (f  )*  =  |  X  f  X  f  =8*g*8  =  &.     Ans. 

243.  Rule  35. — (a)   To  raise  a  whole  number  or  a  deci- 
mal to  any  power,  use  it  as  a  factor  as  many  times  as  there 
are  units  in  the  exponent. 

(b)     To  raise  a  fraction  to  any  power,  raise  both  the  numer- 
ator and  denominator  to  the  power  indicated  by  the  exponent. 


EXAMPLES  FOR  PRACTICE. 

244.      Raise  the  following  to  the  powers  indicated: 

(a)     852.  f  (a)      7,225. 

(*)  (H)4-  .     I  w  m- 

(c)  6.5s.  Ans'  1    (c)     42.25. 

(d)  14*.  I    (d)     38,416. 


ARITHMETIC.  79 

(')  (f)3-  {  (')  If 

(/)  (I)3-  An,     (/)  HI- 

(£•)  (i)3-  1  (g-)  4s- 

(>*)  1.45.  [   (A)  5.37824. 

EVOLUTION. 

245.  Evolution  is  the  reverse  of  involution.      It  is  the 
process  of  finding  the  root  of  a  number  which  is  considered 
as  a  power. 

246.  The  square  root  of  a  number  is  that  number 
which,  when  used  twice  as  a  factor,  produces  the  number. 

Thus,  2  is  the  square  root  of  4,  since  2  X  2,  or  2",  =  4. 

247.  The  cube  root  of  a  number  is  that  number  which, 
when  used  three  times  as  a  factor,  produces  the  number. 

Thus,  3  is  the  cube  root  of  27,  since  3  X  3  X  3,  or  33,  =27. 

248.  The  radical  sign  ^/,  when  placed  before  a  num- 
ber, indicates  that  some  root  of  that  number  is  to  be  found. 

249.  The  index  of  the  root  is  a  small  figure  placed  over 
and  to  the  left  of  the  radical  sign,  to  show  what  root  is  to  be 
found. 

Thus,  1/100  denotes  the  square  root  of  100. 
1/125  denotes  the  cube  root  of  125. 
4/256  denotes  the  fourth  root  of  256,  and  so  on. 

250.  When  the  square  root  is  to  be  extracted,  the  index 
is  generally  omitted.     Thus,  4/100  indicates  the  square  root 
of  100.     Also,  4/225  indicates  the  square  root  of  225. 


SQUARE    ROOT. 

251.  The  largest  number  that  can  be  written  with  one 
figure  is  9,  and  9*  =  81;  the  largest  number  that  can  be 
written  with  two  figures  is  99,  and  99*  =  9,801;  with  three 
figures  999,  and  999*  =  998,001;  with  four  figures  9,999,  and 
9,999s  =  99,980,001,  etc. 

In  each  of  the  above  it  will  be  noticed  that  the  square  of 
the  number  contains  just  twice  as  many  figures  as  the 
number. 

In  order  to  find  the  square  root  of  a  number,  the  first  step 
is  to  find  how  many  figures  there  will  be  in  the  root.  This 


80  ARITHMETIC. 

is  done  by  pointing  off  the  number  \n\.o  periods  of  two  figures 
each,  beginning  at  tJie  rigJit.  The  number  of  periods  will 
indicate  the  number  of  figures  in  the  root. 

Thus,  the  square  root  of  83,740,801  must  contain  4  figures, 
since,  pointing  off  the  periods,  we  get  83'74'08'01,  or  4  periods ; 
consequently,  there  must  be  4  figures  in  the  root.  In  like 
manner,  the  square  root  of  50,625  must  contain  3  figures, 
since  there  are  (5'06'25)  3  periods. 

252.      EXAMPLE.— Find  the  square  root  of  31,505,769. 

root. 

SOLUTION.—          («)       5  31'50'57'69  ( 561 3    Ans. 

(d) 


5 

31'50'57'69(  5613 

5 

(6)  ^5 

100 

(c)   650 

6 

636 

106 

(e)    1457 

6 

1121 

1120 

33669 

1 

33669 

1121 

0 

1 

11220 

3 

11223 

EXPLANATION. — Pointing  off  into  periods  of  two  figures 
each,  it  is  seen  that  there  are  four  figures  in  the  root.  Now, 
find  the  largest  single  number  whose  square  is  less  than  or 
equal  to  31,  the  first  period.  This  is  evidently  5,  since 
6s  =  36,  which  is  greater  than  31.  Write  it  to  the  right,  as  in 
long  division,  and  also  to  the  left,  as  shown  at  (a).  This  is 
the  first  figure  of  the  root.  Now,  multiply  the  5  at  (a)  by 
the  5  in  the  root,  and  write  the  result  under  the  first  period, 
as  sh~wn  at  (6).  Subtract,  and  obtain  6  as  a  remainder. 

Bring  down  the  next  period  50,  and  annex  it  to  the  re- 
mainder 6,  as  shown  at  (c),  which  we  call  the  dividend. 
Add  the  root  already  found  to  the  5  at  (a),  getting  10,  and 
annex  a  cipher  to  this  10,  thus  making  it  100,  which  we  call 
the  trial  divisor.  Divide  the  dividend  (c)  by  the  trial 
divisor  (</),  and  obtain  6,  which  is  probably  the  next  figure 
of  the  root.  Write  6  in  the  root,  as  shown,  and  also  add  it 


ARITHMETIC.  81 

to  100,  the  trial  divisor,  making  it  106.     This  is  called  the 
complete  divisor. 

Multiply  this  by  6,  the  second  figure  in  the  root,  and  sub- 
tract the  result  from  the  dividend  (c).  The  remainder  is  14, 
to  which  annex  the  next  period,  making  it  1,457,  as  shown 
at  (e\  which  we  call  the  new  dividend.  Add  the  second 
figure  of  the  root  to  the  trial  divisor  106,  and  annex  a  cipher, 
thus  getting  1,120.  Dividing  1,457  by  1,120,  we  get  1  as  the 
next  figure  of  the  root.  Adding  this  last  figure  of  the  root 
to  1,120,  multiplying  the  result  by  it,  and  subtracting  from 
1,457,  the  remainder  is  336. 

Annexing  the  next  and  last  period,  69,  the  result  is  33,669. 
Now,  adding  the  last  figure  of  the  root  to  1,121,  and  annexing 
a  cipher  as  before,  the  result  is  11,220.  Dividing  33,669  by 
11,220,  the  result  is  3,  the  fourth  figure  in  the  root.  Adding  it 
to  11,220,  and  multiplying  the  sum  by  it,  the  result  is  33,669. 
Subtracting,  there  is  no  remainder;  hence, 4/31, 505, 769  = 
5,613.  Ans. 

253.  The  square  of  any  number  wholly  decimal  always 
contains  twice  as  many  figures  as  the  number  squared.    For 
example,  .la  =  .01;  .132  =  .0169;  .751'  =  .564001,  etc. 

254.  It  will  also  be  noticed  that  the  number  squared  is 
always  less  than  the  decimal.      Hence,  if  it  be  required  to 
find  the  square  root  of  a  decimal,  and  the  decimal  has  not  an 
even  number  of  figures  in  it,  annex  a  cipher.     The  best  way 
to  determine  the  number  of  figures  in  the  root  of  a  decimal 
is   to  begin  at  the  decimal  point,  and,  going  towards  the 
right,  point  off  the  decimal  into  periods  of  two  figures  each. 
Then,  if  the  last  period  contains  but  one  figure,  annex  a 
cipher. 

255.  EXAMPLE.— What  is  the  square  root  of  .000576  ? 

root 

SOLUTION.—  2  ,00'05'76(.024     Ans. 

2  4 

~40  776 

4  176 

44  ~~0 


ARITHMETIC. 


EXPLANATION. — Beginning  at  the  decimal  point,  and  point- 
ing off  the  number  into  periods  of  two  figures  each,  it  is  seen 
that  the  first  period  is  composed  of  ciphers;  hence,  the  first 
figure  of  the  root  must  be  a  cipher.  The  remaining  portion 
of  the  solution  should  be  perfectly  clear  from  what  has 
preceded. 

256.  If  the  number  is  not  a  perfect  power,  the  root  will 
consist  of  an  interminable  number  of  decimal  places.     The 
result  may  be  carried  to  any  required  number  of  decimal 
places   by   annexing   periods   of  two   ciphers    each   to   the 
number. 

257.  EXAMPLE.— What  is  the  square  root  of  3  ?     Find  the  result 
to  five  decimal  places. 


SOLUTION. —  1 
1 


3.0000'0000'00(1.  73205  + 
1 


Ans. 


189 

1100 
1029 


7100 
6924 

1760000 
1732025 

27975 


20 

7 

7 

"340 
3 

343 
3 

3460 

2 

3462 

2 

346400 
5 

346405 

EXPLANATION.  — Annexing  five  periods  of  two  ciphers  each 
to  the  right  of  the  decimal  point,  the  first  figure  of  the  root 
is  1.  To  get  the  second  figure,  we  find  that,  in  dividing 
200  by  20,  it  is  10.  This  is  evidently  too  large. 

Trying  9,  we  add  9  to  20,  and  multiply  29  by  9,  the  result 
is  261,  a  result  which  is  considerably  larger  than  200;  hence, 


ARITHMETIC.  83 

9  is  too  large.  In  the  same  way  it  is  found  that  8  is  also  too 
large.  Trying  7,  7  times  27  is  189,  a  result  smaller  than 
200;  therefore,  7  is  the  second  figure  of  the  root.  The  next 
two  figures,  3  and  2,  are  easily  found.  The  fifth  figure  in 
the  root  is  a  cipher,  since  the  trial  divisor  34,640  is  greater 
than  the  new  dividend  17,600.  In  a  case  of  this  kind,  we 
annex  another  cipher  to  34,640,  thereby  making  it  346,400, 
and  bring  down  the  next  period,  making  the  17,600, 1,760,000. 
The  next  figure  of  the  root  is  5,  and  as  we  now  have  five 
decimal  places,  we  will  stop. 

The  square  root  of  3  is,  then,  1.73205  -f .      Ans. 

258.      EXAMPLE.— What  is  the  square  root  of  .3  to  five  decimal 
places  ? 

root 

SOLUTION.—      5  .30'00'00'00'00(  .54772+    Ans. 

5  25 

100  500 

4  416 

I04  ~8400 

4  7609 

fb"?0  79100 

7  76629 


1087  247100 

7  219084 


10940  28016 

7 


10947 

7 

109540 

2 

109542 


EXPLANATION. — In  the  above  example,  we  annex  a  cipher 
to  .3,  making  the  first  period  .30,  since  every  period  of  a 
decimal,  as  was  mentioned  before,  must  have  two  figures  in 
it.  The  remainder  of  the  work  should  be  perfectly  clear. 

259.  If  it  is  required  to  find  the  square  root  of  a  mixed 
number,  begin  at  the  decimal  point,  and  point  off  the 


84  ARITHMETIC. 

periods   both  ways.     The  manner  of   finding  the  root  will 
then  be  exactly  the  same  as  in  the  previous  cases. 

26O.      EXAMPLE.— What  is  the  square  root  of  258.2449  ? 
SOLUTION.—  1  2'58.24'49  ( 1 6.07    Ans. 


1 

1 

2'58.24'49( 

1 

16.07 

20 
6 

158 
156 

26 
6 

22449 
22449 

3200 

7 

0 

3207 

EXPLANATION. — In  the  above  example,  since  320  is  greater 
than  224,  we  place  a  cipher  for  the  third  figure  of  the  root, 
and  annex  a  cipher  to  320,  making  it  3,200.  Then,  bringing 
down  the  next  period  49,  7  is  found  to  be  the  fourth  figure 
of  the  root.  Since  there  is  no  remainder,  the  square  root  of 
258.2449  is  16.07.  Ans. 

261.  Proof. —  To  prove  square  root,  square  the  result 
obtained.     If  the  number  is  an  exact  power,  the  square  of  the 
root  will  equal  it;  if  it  is  not  an  exact  power,  the  square  of 
the  root  will  very  nearly  equal  it. 

262.  Rule  36. — (a)  Begin  at  units  place,  and  separate  the 
number  into  periods  of  two  figures  each,  proceeding  from  left 
to  right  with  the  decimal  part,  if  there  is  any. 

(b)  Find  the  greatest  number  whose  square  is  contained  in 
the  first  or  left-hand  period.      Write  this  number  as  the  first 
figure  in   the  root;  also,  write  it  at  the   left  of  the  given 
number. 

Multiply  this  number  at  the  left  by  the  first  figure  of  the 
root,  and  subtract  the  result  from  the  first  period;  then  annex 
the  second  period  to  the  remainder. 

(c)  Add  the  first  figure  of  the  root  to  the  number  in  the 
first  column  on  the  left,  and  annex  a  cipher  to  the  result;  this 
is  the  trial  divisor.     Divide  the  dividend  by  the  trial  divisor 


ARITHMETIC.  85 

for  the  second  figure  in  the  root,  and  add  this  figure  to  the 
trial  divisor  to  form  the  complete  divisor.  Multiply  the  com- 
plete divisor  by  the  second  figure  in  the  roof,  and  subtract  this 
result  from  the  dividend,  (ff  this  result  is  larger  than  the. 
dividend,  a  smaller  number  must  be  tried  for  the  second  fig- 
ure of  the  root.}  Now  bring  down  the  third  period,  and 
annex  it  to  the  last  remainder  for  a  new  dividend.  Add 
the  second  figure  of  the  root  to  the  complete  divisor,  and  annex 
a  cipher  for  a  new  trial  divisor. 

(d)  Continue   in    this   manner   to   the  last  period,  after 
which,  if  any  additional  places  in  the  root  are  required,  bring 
down  cipher  periods,  and  continue  the  operation. 

(e)  If  at  any  time  the  trial  divisor  is  not  contained  in  the 
dividend,  place  a  cipher  in  the  root,  annex  a  cipher  to  the 
trial  divisor,  and  bring  down  another  period. 

(/)  If  the  root  contains  an  interminable  decimal,  and  it 
is  desired  to  terminate  the  operation  at  some  point,  say  the 
fourth  decimal  place,  carry  the  operation  one  place  further, 
and  if  the  fifth  figure  is  5  or  greater,  increase  the  fourth 
figure  by  1  and  omit  the  sign  -J-. 

263.  Short  Method. — If  the  number  whose  root  is  to 
be  extracted  is  not  an  exact  square,  the  root  will  be  an  in- 
terminable decimal.  It  is  then  usual  to  extract  the  root  to 
a  certain  number  of  decimal  places.  In  such  cases,  the  work 
may  be  greatly  shortened  as  follows:  Determine  to  how 
many  decimal  places  the  work  is  to  be  carried,  say  5,  for  ex- 
ample; add  to  this  the  number  of  places  in  the  integral  part 
of  the  root,  say  2,  for  example,  thus  determining  the  num- 
ber of  figures  in  the  root,  in  this  case  5  +  2  —  7.  Divide  this 
number  by  2  and  take  the  next  higher  number.  In  the 
above  case,  we  have  7  -r-  2  =  3£;  hence,  we  take  4,  the  next 
higher  number.  Now  extract  the  root  in  the  usual  manner 
until  the  same  number  of  figures  have  been  obtained  as  was 
expressed  by  the  number  obtained  above,  in  this  case  4. 
Then  form  the  trial  divisor  in  the  usual  manner,  but  omit- 
ting to  annex  the  cipher ;  divide  the  last  remainder  by  the  trial 


86  ARITHMETIC. 

divisor,  as  in  long  division,  obtaining  as  many  figures  of  the 
quotient  as  there  are  remaining  figures  of  the  root,  in  this 
case  7  —  4  =  3.  The  remainder  so  obtained  is  the  remain- 
ing figures  of  the  root. 

Consider  the  example  in  Art.  258.  Here  there  are  5  fig- 
ures in  the  root.  We  therefore  extract  the  root  to  3  places 
in  the  usual  manner,  obtaining  .547  for  the  first  three 
root  figures.  The  next  trial  divisor  is  1,094  (with  the 
cipher  omitted),  and  the  last  remainder  is  791.  Then,  791  -r- 
1,094=  .723,  and  the  next  two  figures  of  the  root  are  72, 
the  whole  root  being  .54772+.  Always  carry  the  division 
one  place  further  than  desired,  and  if  the  last  figure  is  5  or 
greater,  increase  the  preceding  figure  by  1.  This  method 
should  not  be  used  unless  the  root  contains  five  or  more 
figures. 

NOTE.— If  the  last  figure  of  the  root  found  in  the  regular  manner  is  a 
cipher,  carry  the  process  one  place  further  before  dividing  as  described 
above. 


264. 


EXAMPLES  FOR   PRACTICE. 

Find  the  square  root  of 


(a)   186,624. 
(£)   2,050,624. 

(c)  29,855,296. 

(d)  .0116964. 

(e)  198.1369. 
(/)  994,009. 

(£•)  2.375  to  four  decimal  places. 
(K)   1.625  to  three  decimal  places. 
(/)    .3025. 
(J)  .571428. 
(k)   .78125. 


Ans.  * 


(a)  432. 
(d)   1,432. 

(c)  5,464. 

(d)  .1081+. 
(<?)    14.0761. 
(/)  997. 
(ff)  1-5411. 

(A)   1.275. 
(/)   .55. 
(/)  .7559+. 
(k)  .8839. 


CUBE    ROOT. 

265.  In  the  same  manner  as  in  the  case  of  square  root, 
it  can  be  shown  that  the  periods  into  which  a  number  is 
divided,  whose  cube  root  is  to  be  extracted,  must  contain 


ARITHMETIC.  87 

three  figures,  except  that  the  first  or  left-hand  period  of  a 
whole  or  mixed  number  may  contain  one,  two,  or  three 
figures. 


266.      EXAMPLE.— What  is  the  cube  root  of  375, 741, 1 
SOLUTION.— 

(3)  root 

3  7  5'7  4 1'8  5  3'6  9  6  (  7  2  1  6  Ans, 


(1) 
7 
7 


14 

7 

210 


212 
2 


2160 
1 

2161 

1 

2162 

1 

21630 
6 


(2) 
49 
98 

77700 

424 

15124 

428 

1555200 
2161 

1557361 
2162 

155952300 
129816 

156082116 


343 

~32741 

30248 

2493853 
1557361 

936492696 
936492696 


EXPLANATION. —  Write  the  work  in  three  columns  as 
follows :  On  the  right  place  the  number  whose  cube  root  is  to 
be  extracted,  and  point  it  off  into  periods  of  three  figures  each. 
Call  this  column  (3).  Find  the  largest  number  whose  cube 
is  less  than  or  equal  to  the  first  period,  in  this  case  7.  Write 
the  7  on  the  right,  as  shown,  for  the  first  figure  of  the  root, 
and  also  on  the  extreme  left  at  the  head  of  column  (1). 
Multiply  the  7  in  column  (1)  by  the  first  figure  of  the  root 
7,  and  write  the  product  49  at  the  head  of  column  (2). 
Multiply  the  number  in  column  (2)  by  the  first  figure  of  the 
root  7,  and  write  the  product  343  under  the  figures  in  the 


88  ARITHMETIC. 

first  period.  Subtract  and  bring  down  the  next  period,  ob- 
taining 32,741  for  the  dividend.  Add  the  first  figure  of  the 
root  to  the  number  in  column  (1),  obtaining  14,  which  call 
the  first  correction.  Multiply  the  first  correction  by  the 
first  figure  of  the  root,  add  the  product  to  the  number  in 
column  (2),  and  obtain  147.  Add  the  first  figure  of  the 
root  to  the  first  correction,  and  obtain  21,  which  call  the 
second  correction.  Annex  two  ciphers  to  the  number  in 
column  (2),  and  obtain  14,700  for  the  trial  divisor;  also,  annex 
one  cipher  to  the  second  correction,  and  obtain  210.  Divi- 
ding the  dividend  by  the  trial  divisor,  we  obtain  —L- —  =  2  +, 

14, 700 

and  write  the  2  as  the  second  figure  of  the  root.  Add  the 
2  to  the  second  correction,  and  obtain  212,  which,  multiplied 
by  the  second  figure  of  the  root,  and  added  to  the  trial 
divisor,  gives  15,124,  the  complete  divisor.  This  last  result, 
multiplied  by  the  second  figure  of  the  root  and  subtracted 
from  the  dividend,  gives  a  remainder  of  2,493.  Annexing 
the  third  period,  we  obtain  2,493,853  for  the  new  dividend. 
Adding  the  second  figure  of  the  root  to  the  number  in 
column  (1 ),  we  get  214  as  the  new  first  correction ;  this,  multi- 
plied by  the  second  figure  of  the  root  and  added  to  the  trial 
divisor,  gives  15,552.  Adding  the  second  figure  of  the  root 
to  the  first  new  correction  gives  216  as  the  second  new 
correction.  Annexing  two  ciphers  to  the  number  in  column 
(2)  gives  1,555,200,  the  new  trial  divisor.  Annexing  one 
cipher  to  the  second  new  correction  gives  2,160.  Divi- 
ding the  new  dividend  by  the  new  trial  divisor,  we  obtain 

2  493  853 

—  =  1  -)-,  and  write  1  as  the  third  figure  of  the  root. 
l,ooo,*00 

The  remainder  of  the  work  should  be  perfectly  clear  from 
what  has  preceded. 

267.  In  extracting  the  cube  root  of  a  decimal,  proceed 
as  above,  taking  care  that  each  period  contains  three  figures. 
Begin  the  pointing  off  at  the  decimal  point,  going  towards 
the  right.  If  the  last  period  does  not  contain  three  figures, 
annex  ciphers  until  it  does. 


ARITHMETIC.  89 

268.      EXAMPLE.— What  is  the  cube  root  of  .009129329  ? 

root 

SOLUTION.—  2  4  .0091  29'329  (  .209 

2  _8  8 

4  120000  1129329 

2  5481  1129329 

600  125481  0 


EXPLANATION. — Beginning  at  the  decimal  point,  and  point- 
ing off  as  shown,  the  largest  number  whose  cube  is  less  than 
9  is  seen  to  be  2 ;  hence,  2  is  the  first  figure  of  the  root. 
When  finding  the  second  figure,  it  is  seen  that  the  trial  divi- 
sor 1,200  is  greater  than  the  dividend;  hence,  write  a  cipher 
for  the  second  figure  of  the  root;  bring  down  the  next  period 
to  form  the  new  dividend;  annex  two  ciphers  to  the  trial 
divisor  to  form  a  new  trial  divisor;  also,  annex  one  cipher  to 
the  60  in  column  (1).  Dividing  the  new  dividend  by  the  new 

trial  divisor,  we  get    '   '  J'   * J  =  9-+,  and  write  9  as  the  third 
li«U,UUU 

figure  of  the  root.     Complete  the  work  as  before. 
269.     EXAMPLE.  —What  is  the  cube  root  of  78,292.892952  ? 
SOLUTION. —  root 

4  16  78'292.892'952    (42.78 

1  I2,  ii 

8          4800          14292 
_4_  244          10088 

120         5044  4204892 

2          248  3766483 

f22         52T200          438409952 
2  8869          438409952 

124         538069  0 

2  8918 

1260        54698700 

7          102544 

1267        54801244 

7 

1274 

7 

12810 

8 

12818 


90 


ARITHMETIC. 


EXPLANATION. — Since  we  have  a  mixed  number,  begin 
at  the  decimal  point  and  point  off  periods  of  three  figures 
each,  in  both  directions.  The  first  period  contains  but  two 
figures,  and  the  largest  number  whose  cube  is  less  than  78  is 
4;  consequently,  4  is  the  first  figure  of  the  root.  The  re- 
mainder of  the  work  should  be  perfectly  clear.  When  divi- 
ding the  dividend  by  the  trial  divisor  for  the  third  figure  of 
the  root,  the  quotient  was  8  -4-  ;  but,  on  trying  it,  it  was  found 
that  8  was  too  large,  the  complete  divisor  being  considerably 
larger  than  the  trial  divisor.  Therefore,  7  was  used  instead 
of  8. 

27O.      EXAMPLE.— What  is  the  cube  root  of  5  to  five  decimal  places? 


SOLUTION.— 

1 
1 

2 

1 
30 
7 
37 

7 

44 

7 

1         5.000'000'000'000'000(1.70997 
2          1 

300        4000 
259        3913 

559 
308 

87000000 
78443829 

8670000 
45981 

8556171000 
7889992299 

8715981 
46062 

666178701000 
614014317973 

5100 
9 

876204300 
461511 

52164383027 

5109 
9 

876665811 
461592 

5118 
9 

87712740300 
3590839 

51270 
9 

87716331139 

51279 
9 

51288 
9 

512970 

7 

512977 

ARITHMETIC. 


91 


EXPLANATION. — In  the  preceding  example  we  annex  five 
periods  of  ciphers,  of  three  ciphers  each,  to  the  5  for  the 
decimal  part  of  the  root,  placing  the  decimal  point  between 
the  5  and  the  first  cipher.  Since  it  is  easy  to  see  that  the 
next  figure  of  the  root  will  be  5,  we  increase  the  last  figure 
by  1,  obtaining  1.70998  for  the  correct  root  to  5  decimal 
places.  Ans. 


271.      EXAMPLE.— What  is  the  cube  root  of  .5  to  four  decimal 
places  ? 

SOLUTION.— 


7 
7 

14 

7 

210 


219 
9 

228 


2370 
3 

2373 


49 
98. 

14700 
1971 

16671 
2052 

1872300 
7119 

1879419 
7128 

188654700 
166579 

188821279 


.500'000'000'000(.7937  + 
343 

157000 
150039 


6961000 
5638257 

1322743000 
1321748953 

994047 


237 


23790 

7 

23797 

EXPLANATION. — In  the  above  example  we  annex  two 
ciphers  to  the  .5  to  complete  the  first  period,  and  three 
periods  of  three  ciphers  each.  The  cube  root  of  500  is  7 ; 
this  we  write  as  the  first  figure  of  the  root.  The  remainder 
of  the  work  should  be  perfectly  plain  from  the  explanations 
of  the  preceding  examples. 


ARITHMETIC. 


272.     EXAMPLE.— What  is  the  cube  root  of  .05  to  four  decimal 
places  ? 
SOLUTION. — 


90 
6 

~9~6 

6 

102 
6 

10~80 

8 

1088 


9 

18 

2700 
576 

3276 
612 

388800 

8704 

397504 

8768 

40627200 

44176 

40671376 


.0  5  O'O  0  O'O  0  O'O  0  0  ( .3  6  8  4  -+• 

27 

Tsooo 

19656 


3344000 
3180032 

163968000 

162685504 

1282496 


1096 


11040 
4 

11044 

273.  Proof. —  To  prove  cube  root,  cube  the  result  ob- 
tained.    If  the  given  number  is  an  exact  power,  the  cube  of 
the  root  will  equal  it ;  if  not  an  exact  power,  the  cube  of  the 
root  will  very  nearly  equal  it. 

274.  Rule  37. — (a)  A  rrange  the  work  in  three  columns, 
placing  the  number  whose  cube  root  is  to  be  extracted  in  the 
third  or  right-hand  column.     Begin  at  units  place,  and  sep- 
arate the  number  into  periods  of  three  figures  each,  proceed- 
ing from    the   decimal  point   towards   the   right   with   the 
decimal  part,  if  there  is  any. 

(b)  Find  the  greatest  number  whose  cube  is  not  greater 
than  the  number  in  the  first  period.  Write  this  number  as 
the  first  figure  of  the  root ;  also,  write  it  at  the  head  of  the 
first  column.  Multiply  the  number  in  the  first  column  by  the 
first  figure  in  the  root,  and  write  the  result  in  the  second 


ARITHMETIC.  93 

column.  Multiply  the  number  in  the  second  column  by  the 
first  figure  of  the  root ;  subtract  the  product  front  the  first 
period,  and  annex  the  second  period  to  the  remainder  for  a 
new  dividend ;  add  the  first  figure  of  the  root  to  the  number 
in  the  first  column  for  the  first  correction.  Multiply  the 
first  correction  by  the  first  figure  of  the  root,  and  add  the 
product  to  the  number  in  the  second  column.  Add  the  first 
figure  of  the  root  to  the  first  correction  to  form  the  second 
correction.  Annex  one  cipher  to  the  second  correction,  and 
two  ciphers  to  the  last  number  in  the  second  column  ;  the  last 
number  in  the  second  column  is  the  trial  divisor. 

(c)  Divide  the  dividend  by  the  trial  divisor  to  find  the 
second  figure  of  the  root.  Add  the  second  figure  of  the  root 
to  the  number  in  the  first  column,  multiply  the  sum  by  the 
second  figure  of  the  root,  and  add  the  result  to  the  trial  divi- 
sor to  form  the  complete  divisor.  Multiply  the  complete  divi- 
sor by  the  second  figure  of  the  root,  subtract  the  result  from 
the  dividend  in  the  third  column,  and  annex  the  third  period 
to  the  remainder  for  a  new  dividend.  Add  the  second  figure 
of  the  root  to  the  number  in  the  first  column  to  form  the  first 
correction;  multiply  the  first  correction  by  the  second  figure 
of  the  root,  and  add  the  product  to  the  complete  divisor.  Add 
the  second  figure  of  the  root  to  the  first  correction  to  form  the 
second  correction.  Annex  one  cipher  to  the  second  correction, 
and  two  ciphers  to  the  last  number  in  the  second  column  to 
form  the  new  trial  divisor. 

(</)  If  there  are  more  periods  to  be  brought  down,  proceed 
as  before.  If  there  is  a  remainder  after  the  root  of  the  last 
period  has  been  found,  annex  cipher  periods,  and  proceed  as 
before.  TJie  figures  of  the  root  thus  obtained  will  be  decimals. 

(e)  If  the  root  contains  an  interminable  decimal,  and  it  is 
desired  to  terminate  the  operation  at  some  point,  say  the 
fourth  decimal  place,  carry  the  operation  one  place  further, 
and  if  the  fifth  figure  is  5  or  greater,  increase  the  fourth 
figure  by  1  and  omit  the  sign  +. 

Art.  263  can  be  applied  to  cube  root  (or  any  other 
root)  as  well  as  to  square  root.  Thus,  in  the  example, 


94  ARITHMETIC. 

Art.  27O,  there  are  to  be  5  +  1  =  G  figures  in  the  root. 
Extracting  the  root  in  the  usual  manner  to  6  -=-  2  =  3,  say  4 
figures,  we  get  for  the  first  four  figures  1,709.  The  last  re- 
mainder is  8,556,171,  and  the  next  trial  divisor  with  the 
ciphers  omitted  is  8,762,043.  Hence,  the  next  two  figures 
of  the  root  are  8,556,171  -f-  8,762,043  =  .976,  say  ,98.  There- 
fore, the  root  is  1.70998.  _ 

ROOTS    OF    FRACTIONS. 

275.  If  a  given  number  is  in  the  form  of  a  fraction  and 
it  is  required  to  find  some  root  of  it,  the  simplest  and  most 
exact  method  is  to  reduce  the  fraction  to  a  decimal  and 
extract  the  required  root  of  the  decimal.      If,  however,  the 
numerator    and  denominator    of    the   fraction   are    perfect 
powers,  extract  the  required  root  of  each  separately,  and 
write  the  root  of  the  numerator  for  a  new  numerator,  and 
the  root  of  the  denominator  for  a  new  denominator. 

276.  EXAMPLE.—  What  is  the  square  root  of  ^  ? 


277.  EXAMPLE.—  What  is  the  square  root  of  £  ? 
SOLUTION.—  |/1  =  |/.  625  =  .7906. 

278.  EXAMPLE.—  What  is  the  cube  root  of  ||  ? 

SOLUTION.—  A/®L  =  ¥^1  =  f. 
"    64       ^64 

279.  EXAMPLE.—  What  is  the  cube  root  of  £  ? 
SOLUTION.—  Since  i  =  .25,  ty~±  =  ^725  =  .62996  +. 

280.  Rule  38.  —  Extract  tlie  required  root  of  tlie  numer- 
ator and  denominator  separately;  or,  reduce  the  fraction  to  a 
decimal  and  extract  the  root  of  the  decimal. 


EXAMPLES  FOR  PRACTICE. 

281.      Find  the  cube  root  of 


(«)  AV 

(b)  2  to  four  decimal  places. 

(c)  4,180,769,192.462  to  three  decimal  places. 
(<0  T8*V 

to   f        . 

(/)  513,229.783302144  to  three  decimal  places. 


Ans. 


(«)§. 

(b)  1.2599+. 

(c)  1,610.962. 

(d)  .8862+. 

(e)  .72114-. 
(/)  80.064. 


ARITHMETIC.  95 

TO    EXTRACT    OTHER    ROOTS    THAN    THE 

SQUARE    AND    CUBE    ROOTS. 
282.     EXAMPLE.—  What  is  the  fourth  root  of  256  ? 
SOLUTION.— 


Therefore,  J/256  _  4     Ans. 

In  this  example,  4/256,  the  index  is  4,  which  equals  2x2. 
The  root  indicated  by  2  is  the  square  root;  therefore,  the 
square  root  is  extracted  twice. 

283.  EXAMPLE.  —What  is  the  sixth  root  of  64  ? 
SOLUTION.—  |/~64=8. 

^8  =  8. 
Therefore,  ^~64=  2.     Ans. 

In  this  example,  |//64~  the  index  is  6,  which  equals  2x3. 
The  root  indicated  by  3  is  the  cube  root;  therefore,  the 
square  and  cube  roots  are  extracted  in  succession. 

284.  Rule  39.  —  Separate  the  index  of  the  required  root 
into  its  factors  (2's  and  3's),  and  extract  successively  the  roots 
indicated  by  the  several  factors  obtained.  The  final  result  will 
be  the  required  root, 

285.  EXAMPLE.—  What  is  the  sixth  root  of  92,873,580  to  two 
decimal  places  ? 

SOLUTION.—  6  =  3x2.  Hence,  extract  the  cube  root,  and  then 
extract  the  square  root  of  the  result.  ^92,873,580  =  452.8601,  and 
4/452.8601  =  21.*28  +  .  Ans. 

286.  It  matters  not  which  root  is  extracted  first,  but  it 
is  probably  easier  and  more  exact  to  extract  the  cube  root 
first. 


EXAMPLES  FOR  PRACTICE. 

287.   Extract  the 
(a)  Fourth  root  of  100. 

(*)  Fourth  root  of  3,049,800,625.        Ans. 
(c)  Sixth  root  of  9,474,296,896. 


(a)  3. 16227 +. 

(b)  235. 

(c)  46. 


96  ARITHMETIC. 

RATIO. 

288.  Suppose  that  it  is  desired  to  compare  two  num- 
bers, say  20  and  4.     If  we  wish  to  know  how  many  times 
larger  20  is  than  4,  we  divide  20  by  4  and  obtain  5  for  the 
quotient;    thus,  20-^-4=5.      Hence,  we   say    that    20   is   5 
times  as  large  as  4,  i.  e.,  20  contains  5  times  as  many  units 
as  4.     Again,  suppose  we  desire  to  know  what  part  of  20  i? 
4.     We  then  divide  4  by  20  and  obtain  |;  thus,  4  -=-  20  =  -£, 
or  .2.      Hence,  4  is  £  or  .2  of  20.     This  operation  of  compar- 
ing two  numbers  is  termed  finding  the  ratio  of  the  two  num- 
bers.     Ratio,  then,  is  a  comparison.      It  is  evident  that  the 
two  numbers  to  be  compared   must   be   expressed   in   the 
same   unit ;  in   other  words,   the   two  numbers   must  both 
be  abstract  numbers  or  concrete  numbers  of  the  same  kind. 
For  example,  it  would  be  absurd  to  compare  20  horses  with 
4  birds,  or  20  horses   with   4.     Hence,  ratio  may  be  de- 
fined as  a  comparison  between  two  numbers  of  the  same 
kind. 

289.  A  ratio  may  be  expressed  in  three  ways ;  thus,  if 
it  is  desired  to  compare  20  and  4,  and  express  this  compari- 
son as  a  ratio,  it  may  be  done  as  follows :  20  -f-  4 ;  20  :  4,  or 

20 

— .     All  three  are  read  the  ratio  of  20  to  4.     The  ratio  of 

4  to  20   would   be   expressed   thus :    4  -~  20 ;    4  :  20,   or  ^. 

The  first  method  of  expressing  a  ratio,  although  correct,  is 
seldom  or  never  used ;  the  second  form  is  the  one  of tenest 
.  met  with,  while  the  third  is  rapidly  growing  in  favor,  and  is 
likely  to  supersede  the  second.  The  third  form,  called  the 
fractional  form,  is  preferred  by  modern  mathematicians, 
and  possesses  great  advantages  to  students  of  algebra  and 
of  higher  mathematical  subjects.  The  second  form  seems 
to  be  better  adapted  to  arithmetical  subjects,  and  is  one  we 
shall  ordinarily  adopt.  There  is  still  another  way  of  express- 
ing a  ratio,  though  seldom  or  never  used  in  the  case  of  a 
simple  ratio  like  that  given  above.  Instead  of  the  colon,  a 
straight  vertical  line  is  used ;  thus,  20  |  4. 


ARITHMETIC.  97 

290.  The  terms  of  a  ratio  are  the  two  numbers  to  be 
compared ;  thus,  in  the  above  ratio,  20  and  4  are  the  terms. 
When  both  terms  are  considered  together,  they  are  called  a 
couplet ;    when    considered    separately,    the    first   term    is 
called  the  antecedent,  and  the  second  term  the  conse- 
quent.    Thus,  in  the  ratio  20  :  4,  20  and  4  form  a  couplet, 
and  20  is  the  antecedent,  and  4  the  consequent. 

291.  A  ratio  may  be  direct  or  inverse.     The  direct 
ratio  of  20  to  4  is  20  :  4,  while  the  inverse  ratio  of  20  to  4  is 
4  :  20.     The  direct  ratio  of  4  to  20  is  4  :  20,  and  the  inverse 
ratio   is   20  :  4.     An    inverse   ratio   is   sometimes   called    a 
reciprocal    ratio.      The    reciprocal    of    a    number    is    1 

divided  by  the  number.     Thus,  the  reciprocal  of  17  is  — ; 

of  f  is  1  4-f  =  f;  i.e.,  the  reciprocal  of  a  fraction  is  the 
fraction  inverted.  Hence,  the  inverse  ratio  of  20  to  4  may 

be  expressed  as  4  :  20,  or  as  —  :  -.  Both  have  equal  values; 
for,  4^20=},  andi4=lxi  =  J. 

292.  The   term   vary  implies  a  ratio.     When  we  say 
that  two  numbers   vary  as  some  other  two  numbers,  we 
mean  that  the  ratio  between  the  first  two  numbers  is  the 
same  as  the  ratio  between  the  other  two  numbers. 

293.  The  value  of  a  ratio  is  the  result  obtained  by 
performing  the  division  indicated.     Thus,  the  value  of  the 
ratio  20:4  is  5,  it  is  the  quotient  obtained  by  dividing  the 
antecedent  by  the  consequent. 

294.  By  expressing  the  ratio  in  the  fractional  form,  for 
example,  the  ratio  of  20  to  4  as  —-,  it  is  easy  to  see,   from 

the  laws  of  fractions,  that  if  both  terms  be  multiplied,  or 
both  divided  by  the  same  number,  it  will  not  alter  the  value 
of  the  ratio.  Thus, 

20  _  20  X  5  _  100         ,  20      204-4  =  5 
4  ~  4  X  5    "    20  ;  4        44-4         T 


ARITHMETIC. 


295.  It  is  also  evident,  from  the  laws  of  fractions,  that 
multiplying  the  antecedent  or  dividing  the  consequent  mul- 
tiplies the  ratio;  and  dividing  the  antecedent  or  multiplying 
the  consequent  divides  the  ratio. 

296.  When  a  ratio  is  expressed  in  words,  as  the  ratio  of 
20  to  4,  the  first  number  named  is  always  regarded  as  the 
antecedent    and    the   second    as   the    consequent,    without 
regard  to  whether  the  ratio  itself  is  direct  or  inverse.     When 
not  otherwise  specified,  all  ratios  are  understood  to  be  direct. 
To  express  an  inverse  ratio,  the  simplest  way  of  doing  it  is 
to  express  it  as  if  it  were  a  direct  ratio,  with  the  first  num- 
ber named  as  the  antecedent,  and  then  transpose  the  ante- 
cedent to  the  place  occupied  by   the  consequent  and  the 
consequent  to  the  place  occupied  by  the  antecedent;  or  if 
expressed  in  the  fractional  form,  invert  the  fraction.     Thus, 
to  express  the  inverse  ratio  of  20  to  4,  first  write  it  20:  4,  and 

20 

then,  transposing  the  terms,  as  4  :  20;  or  as  — ,  and  then  in- 
verting as  — .     Or,  the  reciprocals  of  the  numbers  may  be 


taken,  as  explained  above, 
pose  its  terms. 


To  Invert  a  ratio  is  to  trans- 


297. 

(a) 


(d) 
w 


EXAMPLES   FOR   PRACTICE. 

What  is  the  value  of  the  ratio  of 
98  to  49  ? 
$45  to  $9  ? 
6itof? 
3.5  to  4.5  ? 

The  inverse  ratio  of  76  to  19  ? 
The  inverse  ratio  of  49  to  98  ? 
The  inverse  ratio  of  18  to  24? 
The  inverse  ratio  of  9  to  15  ? 
The  ratio  of  10  to  3,  multiplied  by  3  ? 
The  ratio  of  35  to  49,  multiplied  by  7  ? 
The  ratio  of  18  to  64,  divided  by  9  ? 
The  ratio  of  14  to  28,  divided  by  5  ? 


Ans. 


'  (a) 

2. 

(i>) 

5. 

(c) 

12i. 

(d) 

.77J. 

W 

J. 

(f) 

2. 

(*•> 

H- 

(K) 

If- 

CO 

10. 

(J) 

5. 

(i) 

&• 

(^) 

rV- 

298.     Instead  of  expressing  the  value  of  a  ratio  by  a 
single  number,  as  above,  it  is  customary  to  express  it  by 


ARITHMETIC.  99 

means  of  another  ratio  in  which  the  consequent  is  1.  Thus, 
suppose  that  it  is  desired  to  find  the  ratio  of  the  weights  of 
two  pieces  of  iron,  one  weighing  45  pounds  and  the  other 
weighing  30  pounds.  The  ratio  of  the  heavier  to  the  lighter 
is  then  45  :  30,  an  inconvenient  expression.  Using  the  frac- 

tional form,  we  have   —  .      Dividing  both  terms  by  30,  the 
ou 

consequent,  we  obtain  -p-  or  1£  :  1.  This  is  the  same  result 
as  obtained  above,  for  1£  -=-  1  =  l£,  and  45  -=-  30  =  1|. 

299.  A  ratio  may  be  squared,  cubed,  or  raised  to  any 
power,  or  any  root  of  it  may  be  taken.  Thus,  if  the  ratio  of 
two  numbers  is  105  :  G3,  and  it  is  desired  to  cube  this  ratio, 
the  cube  may  be  expressed  as  1053  :  633.  That  this  is  correct 
is  readily  seen  ;  for,  expressing  the  ratio  in  the  fractional 

105  .   /105\3     1053 

form,  it  becomes  -  ,  and  the  cube  is  (  --  I  =  —  ^  =  lOo   :  63  . 
63  '  \  63  /        633 

Also,  if  it  is  desired  to  extract  the  cube  root  of  the  ratio 
1053  :  633,  it  may  be  done  by  simply  dividing  the  exponents 
by  3,  obtaining  105  :  63.  This  may  be  proved  in  the  same 
way  as  in  the  case  of  cubing  the  ratio.  Thus,  105'  :  63s  = 


3OO.     Since    -=  it  follows  that  105s  :  63s  = 


=  (|Y,  it 


53  :  33  (this  expression  is  read  :  the  ratio  of  105  cubed  to  G3 
cubed  equals  the  ratio  of  5  cubed  to  3  cubed),  it  follows 
that  the  antecedent  and  consequent  may  always  be  multi- 
plied or  divided  by  the  same  number,  irrespective  of  any 
indicated  powers  or  roots,  without  altering  the  value  of  the 
ratio.  Thus,  24"  :  18J  =  42  :  3".  For,  performing  the  opera- 
tions indicated  by  the  exponents,  241  =  576  and  18'  =  324. 
Hence,  576  :  324  =  1J  or  1$  :  1.  Also,  4'  =  16  and  3*  =  9; 
hence,  16  :  9  =  IJor  1J  .:  1,  the  same  result  as  before.  Also, 

,      24'       /24V  _  /4  V  _  4'  _  , 

M    '  18   =  185  =  W  -  \3  )  -   31  -  4   '  3  ' 


100  ARITHMETIC. 

The  statement  may  be  proved  for  roots  in  the  same  man- 
ner. Thus,  f24~3  :  tflS3  =  %~£  :  4*/3l  For  the  ^sT'1  =  24 
and  ^IS3  =  18;  and,  24  :  18  =  \\  or  1£  :  1-  Also,  fiT*  =  4 
and  $W  =  3;  4  :  3  =  1J  or  1£  :  1. 

NOTE. — If  the  numbers  composing  the  antecedent  and  consequent 
have  different  exponents,  or  if  different  roots  of  those  numbers  are 
indicated,  the  operations  described  in  Art.  SOOcannotbe  performed. 
This  is  evident ;  for,  consider  the  ratio  42  :  83.  When  expressed  in  the 

fractional  form,  it  becomes  — ,  which  can  not  be  expressed  either  as  (  g-  j 

(4  \3 
-Q  j  ,  and,  hence,  can  not  be  reduced  as  described  above. 

PROPORTION. 

301.  Proportion  is  an  equality  of  ratios,  the  equality 
being  indicated  by  the  double  colon  ( : : )  or  by  the  sign  of 
equality  (  =  ).     Thus,  to  write  in  the  form  of  "a  proportion 
the  two  equal  ratios,  8 :  4  and  6 :  3,  which  both  have  the  same 
value  2,  we  may  employ  one  of  the  three  following  forms: 

8  :  4  ::  G  :  3          (1) 
8  :  4  =  G  :  3  (2) 

H       ^ 

302.  The  first  form  is  the  one  most  extensively  used, 
by  reason  of  its  having  been  exclusively  employed  in  all  the 
older  works  on  mathematics.     The  second  and  third  forms 
are  being  adopted  by  all  modern  writers  on  mathematical 
subjects,  and,  in  time,  will  probably  entirely  supersede  the 
first  form.     In  this  paper  we  shall  adopt  the  second  form, 
unless  some  statement  can  be  made  clearer  by  using  the 
third  form.' 

303.  A  proportion  may  be  read  in  two  ways.     The  old 
way  to  read  the  above  proportion  was:  8  is  to  If.  as  6  is  to  3  ; 
the  new  way  is :  the  ratio  of  8  to  4  equals  the  ratio  of  6  to  3. 
The  student  may  read  it  either  way,  but  we  recommend  the 
latter. 

304.  Each  ratio  of  a  proportion   is  termed  a  couplet. 
In  the  above  proportion,  8 :  4  is  a  couplet,  and  so  is  6 :  3. 


ARITHMETIC.  101 

305.  The  numbers  forming  the  proportion  are  called 
terms ;  and  they  are  numbered  consecutively  from  left  to 

right,  thus  : 

first  second  third  fourth 
8:4=6:3 

Hence,  in  any  proportion,  the  ratio  of  the  first  term  to  the 
second  term  equals  the  ratio  of  the  third  term  to  the  fourth 
term. 

306.  The   first  and  fourth  terms  of   a  proportion  are 
called  the  extremes,  and  the  second  and  third  terms  the 
means.     Thus,  in  the  foregoing  proportion,  8  and  3  are  the 
extremes  and  4  and  6  are  the  means. 

307.  A  direct  proportion  is  one  in  which  both  coup- 
lets are  direct  ratios. 

308.  An  inverse  proportion  is  one  which  requires 
one  of   the   couplets  to   be  expressed  as  an  inverse  ratio. 
Thus,  8  is  to  4  inversely  as  3  is  to  6  must  be  written  8:4  = 
6  :  3 ;  i.  e. ,  the  second  ratio  (couplet)  must  be  inverted. 

309.  Proportion  forms  one  of  the  most  useful  sections 
of  arithmetic.     In   our   grandfathers'   arithmetics,   it   was 
called  'l  The  rule  of  three." 

310.  Rule  4O. — In  any  proportion,  the  product  of  the 
extremes  equals  the  product  of  t/ie  means. 

Thus,  in  the  proportion, 

17  :  51  =  14  :  42. 
17  X  42  =  51  X  14,  since  both  products  equal  714. 

311.  Rule  41. —  The  product  of  the  extremes  divided 
by  either  mean  gives  the  other  mean. 

EXAMPLE. — What  is  the  third  term  of  the  proportion  17  :  51  =    :  42  ? 
SOLUTION.— Apply  ing  the  rule,  17  X  42  =  714,  and  714-*- 51  =  14.  Ans. 

312.  Rule  42.—  The  product  of  the  means  divided  by 
either  extreme  gives  the  other  extreme. 

EXAMPLE.— What  is  the  first  term  of  the  proportion     :  51  =  14  :  42  ? 
SOLUTION.— Applying  the  rule,  51  X  14  =  714,  and  714  -H  42  =  17.  Ans. 


102  ARITHMETIC. 

313.  When  stating  a  proportion  in  which  one  of  the 
terms  is  unknown,  represent  the   missing  term  by  a  letter, 
as  x.     Thus,  the  last  example  would  be  written, 

x  :  51  =  14  :  42 

and  for  the  value  of  x  we  have  x  =  — — —  =  17. 

314.  If  the  same  (addition  and  subtraction  excepted) 
operations  be  performed  upon  all  of  the  terms  of  a  propor- 
tion, the   proportion    is   not   thereby    destroyed.      In  other 
words,  if  all  of  the  terms  of  a  proportion  be  (1)  multiplied 
or  (2)  divided  by  the  same  number;  (3)  if  all  the  terms  be 
raised  to  the  same  power;  if  (4)  the  same  root  of  all  the 
terms  be  taken,  or  (5)  if  both  couplets  be  inverted,  the  pro- 
portion still  holds.     We  will  prove  these  statements  by  a 
numerical  example,  and  the  student  can  satisfy  himself  by 
other  similar  ones.     The  fractional  form  will  be  used,  as  it 
is  better  suited  to  the  purpose.     Consider  the  proportion 
8:4=6:3.      Expressing  it  in  the  third  form,  it  becomes 

—  =  -,  What  we  are  to  prove  is  that,  if  any  of  the  five 
4  o 

operations  enumerated  above  be  performed  upon  all  of  the 
terms  of  this  proportion,  the  first  fraction  will  still  equal 
the  second  fraction. 

O     y    IV 

1.  Multiplying  all  the  terms  by  any  number,  say  7,  j = 

6  X  7  56        42  56  ,    42 

=  33TT  °r  2~8  =  21-  N°W  28  6Vldently  6qUals  21'  SmCC 
the  value  of  either  ratio  is  2,  and  the  same  is  true  of  the 

original  proportion. 

q    .    iy 

2.  Dividing  all  the  terms  by  any  number,  say  7,  = 

1^];  or  |  =  i.     But  |->  |  =  2,  and  f  +  f  =  S  also,  the 

same  as  in  the  original  proportion. 

3.  Raising  all  the  terms  to  the  same  power,  say  the  cube, 

|]  =  ||.     This  is  evidently  true,  since  |[  =  (|Y=  23  =  8, 


ARITHMETIC.  103 

4.     Extracting  the  same  root  of  all  the  terms,  say  the  cube 

root,  -^=  =  57=.     It   is   evident   that  this  is  likewise   true, 


^8         8/8 
=  V  4  = 


4  =        >  and  jfl  =       "3  =         alsa 

4      3 
5.     Inverting  both  couplets,  5-=—,  which   is  true,  since 

o       b 
both  equal  \. 

315.  If  both  terms  of  either  couplet  be  multiplied  or 
both  divided  by  the  same  number,  the  proportion  is  not  de- 
stroyed.    This  should  be  evident  from  the  preceding  article, 
and  also  from  Art.  294.     Hence,  in  any  proportion,  equal 
factors  may  be  canceled  from  the  terms  of  a  couplet,  before 
applying  rule  41  or  42.     Thus,  the  proportion  45:9  =  x\ 
7.  1,  we  may  divide  both  terms  of  the  first  couplet  by  9  (that 
is,  cancel  9  from  both  terms),  obtaining  5:  1  =  x\  7.1,  whence 
x  =  1.  1  x  5  -T-  I  =  35.5.     (See  note  in  Art.  3OO.) 

316.  The  principle  of  all  calculations  in  proportion  is 
this  :     Three  of  the  terms  are  always  given,  and  the  remain- 
ing one  is  to  be  found. 

317.  EXAMPLE.  —  If  4  men  can  earn  $25  in  one  week,  how  much 
can  12  men  earn  in  the  same  time  ? 

SOLUTION.  —  The  required  term  must  bear  the  same  relation  to  the 
given  term  of  the  same  kind  as  one  of  the  remaining  terms  bears  to 
the  other  remaining  term.  We  can  then  form  a  proportion  by  which 
the  required  term  may  be  found. 

The  first  question  the  student  must  ask  himself  in  every  calculation 
by  proportion  is  : 

"What  is  it  I  want  to  find  ?" 

In  this  case  it  is  dollars.  We  have  two  sets  of  men,  one  set  earning 
$25,  and  we  want  to  know  how  many  dollars  the  other  set  earns.  It  is 
evident  that  the  amount  12  men  earn  bears  the  same  relation  to  the 
amottnt  that  4  men  earn  as  12  men  bears  to  4  men.  Hence,  we  have  the 
proportion,  the  amount  12  men  earn  is  to  $25  as  12  men  is  to  4  men; 
or,  since  either  extreme  equals  the  product  of  the  means  divided  by  the 
other  extreme,  we  have 

The  amount  12  men  earn  :  $25  =  12  men  :  4  men, 

CJOPJ  v  12 
or  the  amount  12  men  earn  =  —      —  =  $75.     Ans. 


104  ARITHMETIC. 

Since  it  matters  not  which  place  x  or  the  required  term  occupies,  the 
problem  could  be  stated  as  any  of  the  following  forms,  the  value  of  x 
being  the  same  in  each  : 

(a)  §25  :  the  amount    12  men    earn  =  4  men  :  12  men  ;    or  the 

amount   12  men  earn  =      °* — ,  or  $75,   since  either   mean  equals 

the  product  of  the  extremes  divided  by  the  other  mean. 

(b)  4  men  :  12  men  =  $25  :  the    amount   12  men  earn ;    or    the 

amount  that  12  men  earn  = ^ ,  or    $75,    since  either  extreme 

equals  the  product  of  the  means  divided  by  the  other  extreme. 

(c)  12    men  :  4    men  =  the    amount    12  men    earn  :  $25 ;    or   the 

QQPj  vx  -JO 

amount  that  12  men  earn  = ,  or  §75.  since  either  mean  equals 

the  product  of  the  extremes  divided  by  the  other  mean. 

318.  If  the  proportion  is  an  inverse  one,  first  form  it 
as  though  it  were  a  direct  proportion,  and  then  invert  one 
of  the  couplets. 


EXAMPLES  FOR   PRACTICE. 

319.      Find  the  value  of  x  in  each  of  the  following: 


(a)  §16  :  §64  ::  x  :  $4. 

(l>)  x  :  85  ::  10  :  17. 

(c)  24  :  x  ::  15  :  40. 

(d)  18  :  94  ::  2  :  x.  Ans. 

(e)  $75  :  §100  =  x  :  100. 
(/)  15pwt.  :jr=21:10. 
(g)  x  :  75  yd.  =  §15  :  §5. 


(a)  jr  =  $l. 

(6)  x  =  50. 

(c)  *  =  64. 

(d)  x=  lOf. 

(e)  x  =  75. 
(/)  *=7}pwt. 
(g)  .r  =  225yd. 


1.  If  75  pounds  of  lead  cost  $2.10,  what  would  125  pounds  cost  at 
the  same  rate  ?  Ans.  §3.50. 

2.  If  A  does  a  piece  of  work  in  4  days  and  B  does  it  in  7  days,  how 
long  will  it  take  A  to  do  what  B  does  in  63  days  ?  Ans.  36  days. 

3.  The  circumferences  of  any  two  circles  are  to  each  other  as  their 
diameters.     If  the  circumference  of  a  circle  7  inches  in  diameter  is 
22  inches,  what  will   be  the  circumference  of  a  circle  31   inches  in 
diameter  ?  Ans.  97f  inches. 


INVERSE    PROPORTION. 

32O.  In  Art.  3O8,  an  inverse  proportion  was  defined 
as  one  which  required  one  of  the  couplets  to  be  expressed  as 
an  inverse  ratio.  Sometimes  the  word  inverse  occurs  in  the 


ARITHMETIC.  10o 

statement  of  the  example  ;  in  such  cases  the  proportion 
can  be  written  directly,  merely  inverting  one  of  the  coup- 
lets. But  it  frequently  happens  that  oniy  by  carefully 
studying  the  conditions  of  the  example  can  it  be  ascertained 
whether  the  proportion  is  direct  or  inverse.  When  in  doubt, 
the  student  can  always  satisfy  himself  as  to  whether  the 
proportion  is  direct  or  inverse  by  first  ascertaining  what  is 
required,  and  stating  the  proportion  as  a  direct  proportion. 
Then,  in  order  that  the  proportion  may  be  true,  if  the  first 
term  is  smaller  than  the  second  term,  the  third  term  must 
be  smaller  than  the  fourth  ;  or  if  the  first  term  is  larger 
than  the  second  term,  the  third  term  must  be  larger  than 
the  fourth  term.  Keeping  this  in  mind,  the  student  can 
always  tell  whether  the  required  term  will  be  larger  or 
smaller  than  the  other  term  of  the  couplet  to  which  the  re- 
quired term  belongs.  Having  determined  this,  the  student 
then  refers  to  the  example,  and  ascertains  from  its  condi- 
tions whether  the  required  term  is  to  be  larger  or  smaller 
than  the  other  term  of  the  same  kind.  If  the  two  determi- 
nations agree,  the  proportion  is  direct;  otherwise,  it  is 
inverse,  and  one  of  the  couplets  must  be  inverted. 

321 .      EXAMPLE.  —If  A's  rate  of  doing  work  is  to  B's  as  5  :  7,  and 

A  does  a  piece  of  work  in  42  days,  in  what  time  will  B  do  it  ? 

SOLUTION. — The  required  term  is  the  number  of  days  it  will  take 
B  to  do  the  work.  Hence,  stating  as  a  direct  proportion. 

5  :  7  =  42  :  x. 

Now,  since  7  is  greater  than  5,  x  will  be  greater  than  42.  But,  referring 
to  the  statement  of  the  example,  it  is  easy  to  see  that  B  works  faster 
than  A;  hence  it  will  take  B  a  less  number  of  days  to  do  the  work  than 
A.  Therefore,  the  proportion  is  an  inverse  one,  and  should  be  stated 

5  :  7  =  .r  :  42, 
from  which  x  =  rl*^  =  30 days.   Ans. 

Had  the  example  been  stated  thus:  The  time  that  A  requires  to  do  a 
piece  of  work  is  to  the  time  that  B  requires,  as  5  :  7;  A  can  do  it  in  42 
days,  in  what  time  can  B  do  it  ?  it  is  evident  that  it  would  take  B  a 
longer  time  to  do  the  work  than  it  would  A;  hence,  x  would  be  greater 

7  x  42 
than  42,  and  the  proportion  would  be  direct,  the  value  of  x  being  — 

=  58.8  days. 


106  ARITHMETIC. 

EXAMPLES  FOR   PRACTICE. 

322.  Solve  the  following: 

1.  If  a  pump  which  discharges  4  gal.  of  water  per  min.  can  fill  a  tank 
in  20  hr.,  how  long  will  it  take  a  pump  discharging  12  gal.  per  min.  to 
fill  it?  Ans.  6|-hr. 

2.  The  circular  seam  of  a  boiler  requires  50  rivets  when  the  pitch 
is  2£  in. ;   how  many  would  be  required  if  the  pitch  were  3|  in.  ? 

Ans.  40. 

3.  The  spring  hangers  on  a  certain  locomotive  are  2^  in.  wide  and 
|  in.  thick ;   those  on  another  engine  are  of  same  sectional  area,  but 
are  3  in.  wide ;  how  thick  are  they  ?  Ans.  £  in. 

4.  A  locomotive  with  driving  wheels  16  ft.  in  circumference  runs  a 
certain  distance  in  5,000  revolutions;  how  many  revolutions  would  it 
make  in  going  the  same  distance,  if  the  wheels  were  22  ft.  in  circum- 
ference (no  allowance  for  slip  being  made  in  either  case)  ? 

Ans.  3, 636^  rev. 

POWERS  AND  ROOTS  IN  PROPORTION. 

323.  It  was  stated  in  Art.  299  that  a  ratio  could  be 
raised  to  any  power  or  any  root  of  it  might  be  taken.     A 
proportion  is  frequently  stated  in  such  a  manner  that  one  of 
the  couplets  must  be  raised  to  some  power  or  some  root  of 
it  must  be  taken.      In  all  such  cases,  both  terms  of  the  coup- 
let so  affected  must  be  raised  to  the  same  power  or  the  same 
root  of  both  terms  must  be  taken. 

324.  EXAMPLE. — Knowing  that  the  weight  of  a  sphere  varies 
as  the  cube  of  its  diameter,  what  is  the  weight  of  a  sphere  6  inches  in 
diameter  if  a  sphere  8  inches  in  diameter  of  the  same  material  weighs 
180  pounds  ? 

SOLUTION. — This  is  evidently  a  direct  proportion.     Hence,  we  write 

63  :  83  =  x  :  180. 
Dividing  both  terms  of  the  first  couplet  by  23  (see  Art.  3OO), 

33  :  4s  =  x  :  180,  or  27  :  64  =  x  :  180  ; 

27  y  1 80 
whence,  x  = ^ —  =  75^£  pounds.     Ans. 

EXAMPLE. — A  sphere  8  inches  in  diameter  weighs  180  pounds  ;  what 
is  the  diameter  of  another  sphere  of  the  same  material  which  weighs 
75H  pounds  ? 

SOLUTION.— Since  the  weights  of  any  two  spheres  are  to  each  other 
as  the  cubes  of  their  diameters,  we  have  the  proportion 

180  :  75}f  =  83  :  x*. 


ARITHMETIC.  107 

x,  the  required  term,  must  be  cubed,  because  the  other  term  of  the 
couplet  is  cubed  (see  Art.  323).     But,  83  =  512;  hence,  ' 

7^1  5   v  PCI  O 

180  :  75}  f  =  512  :  x\  or  x*  =  =  216  ; 


whence,  x  =   y'aiG  =  6  inches.     Ans. 

325.  Since  taking  the  same  root  of  all  of  the  terms  of  a 
proportion  does  not  change  its  value  (Art.  314),  the  above 
example  might  have  been  solved  by  extracting  the  cube  root 
of  all  of  the  numbers,  thus  obtaining  $/~ISQ  ;  ^7515  _ 


8  x  f  =  6  inches.     The  process,  however,  is  longer  and  is 
not  so  direct,  and  the  first  method  is  to  be  preferred. 

326.  If  two  cylinders  have  equal  volumes,  but  different 
diameters,  the  diameters  are  to  each  other  inversely  as  the 
square  roots  of  their  lengths.  Hence,  if  it  is  desired  to  find 
the  diameter  of  a  cylinder  that  is  to  be  15  inches  long,  and 
which  shall  have  the  same  volume  as  one  that  is  9  inches  in 
diameter  and  12  inches  long,  we  write  the  proportion 
9  :  *=yi5  :  4/12. 

Since  neither  12  nor  15  are  perfect  squares,  we  square  all 
of  the  terms  (Arts.  325  and  314)  and  obtain 

81  :  x*  =  15  :  12;  whence  x*  =  81  X  12  =  64.8, 

1  U 

and  x  =  4/64.8  =  8.05  inches  =  diameter  of  15-inch  cylinder. 


EXAMPLES  FOR  PRACTICE. 

327.      Solve  the  following  examples: 

1.  The  intensity  of  light  varies  inversely  as  the  square  of  the  dis- 
tance from  the  source  of  light.     If  a  gas  jet  illuminates  an  object  80 
feet  away  with  a  certain  distinctness,  how  much  brighter  will   the 
object  be  at  a  distance  of  20  feet  ?  Ans.  2±  times  as  bright. 

2.  In  the  last  example,  suppose  that  the  object  had  been  40  feet 
from  the  gas  jet;  how  bright  would  it  have  been  compared  with  its 
brightness  at  30  feet  from  the  gas  jet  ?  Ans.  -fg  as  bright. 

3.  When  comparing  one  light  with  another,  the  intensities  of  their 
illuminating  powers  vary  as  the  squares  of  their  distances  from  the 


108  ARITHMETIC. 

source.  If  a  man  can  just  distinguish  the  time  indicated  by  his  watch, 
50  feet  from  a  certain  light,  at  what  distance  could  he  distinguish  the 
time  from  a  light  3  times  as  powerful  ?  Ans.  86.64-  feet. 

4.  The  quantity  of  air  flowing  through  a  mine  varies  directly  as 
the  square  root  of  the  pressure.     If  60,000  cubic  feet  of  air  flow  per 
minute  when  the  pressure  is  2.8  pounds  per  square  foot,  how  much  will 
flow  when  the  pressure  is  3.6  pounds  per  square  foot  ? 

Ans.  68,034  cu.  ft.  per  min.,  nearly. 

5.  In  the  last  example,  suppose  that  70,000  cubic  feet  per  minute 
had  been  required;   what  would  be  the  pressure  necessary  for  this 
quantity  ?  Ans.  3.81+  Ib.  per  sq.  ft. 


CAUSES    AND    EFFECTS. 

328.  Many  examples  in  proportion  may  be  more  easily 
solved  by  using  the  principle  of    cause  and  effect.     That 
which  may  be  regarded  as  producing  a  change  or  alteration 
in  something,  or  as  accomplishing  something,  may  be  called 
a  cause,  and  the  change,  or  alteration,  or   thing   accom- 
plished, is  the  effect. 

329.  Like  causes  produce  like  effects.     Hence,  when  two 
causes  of   the  same  kind  produce  two  effects  of  the  same 
kind,  the  ratio  of  the  causes  equals  the  ratio  of  the  effects; 
in  other  words,  the  first  cause  is  to  the  second  cause  as  the 
first  effect  is  to  the  second  effect.     Thus,  in  the  question,  if 
3  men  can  lift  1,400  pounds,  how  many  pounds  can   7   men 
lift  ?  we  call  3  men  and  7  men  the  causes  (since  they  ac- 
complish something,    viz.,    the  lifting  of  the  weight),    the 
number  of  pounds  lifted,  viz.,  1,400  pounds  and  x  pounds, 
are  the  effects.     If  we  call  3  men  the   first   cause,    1,400 
pounds  is  the  first  effect ;  7  men  is  the  second  cause,  and  x 
pounds  is  the  second  effect.     Hence,  we  may  write 

1st  cause      2d  cause          1st  effect      2 d  effect 

3:7       —       1,400      :      x, 

7  V  1  4-00 

whence  x  =  —          —— 3, 266f  pounds. 
<j 

330.  The  principle  of  cause  and  effect  is  extremely  use- 
ful in  the  solution  of  examples  in  compound  proportion,  as 
we  shall  now  show. 


ARITHMETIC.  109 

COMPOUND    PROPORTION. 

331.  All  the  cases  of  proportion  so  far  considered  have 
been  cases  of  simple  proportion;  i.  e.,  each  term  has 
been  composed  of  but  one  number.  There  are  many  cases, 
however,  in  which  two  or  all  of  the  terms  have  more  than 
one  number  in  them;  all  such  cases  belong  to  compound 
proportion.  In  all  examples  in  compound  proportion, 
both  causes  or  both  effects  or  all  four  consist  of  more  than 
two  numbers.  We  will  illustrate  this  by  an 

EXAMPLE.—  If  40  men  earn  $1,280  in  16  days,  how  much  will  36  men 
earn  in  31  days  ? 

SOLUTION.  —  Since  40  men  earn  something,  40  men  is  a  cause,  and 
since  they  take  16  days  in  which  to  earn  something,  16  days  is  also  a 
cause.  For  the  same  reason,  36  men  and  31  days  are  also  causes.  .  The 
effects,  that  which  is  earned,  are  1,280  dollars  and  x  dollars.  Then,  40 
men  and  16  days  make  up  the  first  cause,  and  36  men  and  31  days  make 
up  the  second  cause.  $1,280  is  the  first  effect  and  %x  is  the  second 
effect.  Hence,  we  write 

1st  cause    2d  cause        1st  effect    2d  effect 

40  36  1  280    • 

16        '        31  ll280    ' 

Now,  instead  of  using  the  colon  to  express  the  ratio,  we  shall  use  the 
vertical  line  (see  Art.  289),  and  the  above  becomes 
40      I      36    _   1  2^      I 
16      |      31    -'   lt280      |    *' 

In  the  last  expression,  the  product  of  all  of  the  numbers  included 
between  the  vertical  lines  must  equal  the  product  of  all  the  numbers 
without  them;  i.  e.,  36  x  31  X  1,260  =  40  X  16  X  x. 

2 


332.  The  above  might  have  been  solved  by  canceling 
factors  of  the  numbers  in  the  original  proportion.  For  if 
any  number  within  the  lines  has  a  factor  common  to  any 
number  without  the  lines,  that  factor  may  be  canceled  from 
both  numbers.  Thus,  16  is  contained  in 

2 
36  ffl 


1,280,  80  times.     Cancel  16  and  1,280,  and  write  80  above 
1,280.     40  is  contained  in  80,  2   cimes.     Cancel  40  and  80, 


110 


ARITHMETIC. 


and  write  2  above  80.  Now,  since  there  are  no  more  num- 
bers that  can  be  canceled,  x  =  36  X  31  X  2  =  $2,232,  he 
same  result  as  was  obtained  in  the  last  article. 

333.  Rule  43. —  Write  all  the  numbers  forming  the  first 
cause  in  a  vertical  column,  and  draw  a  vertical  line  ;  on  the 
other  side  of  this  line  write  in  a  vertical  column  all  of  the 
numbers   forming  the  second    cause.      Write    the    sign    of 
equality  to  the  right  of  the  second  column,  and  on  the  right 
of  this  form  a  third  column  of  the  numbers  composing  the 
first  effect,  drawing  a  vertical  line  to  the  right;  on  the  other 
side  of  this  line,  write,  for  a  fourth  column,   the  numbers 
composing  the  second  effect.      There  must  be  as  many  numbers 
in  the  second  cause  as  in  the  first  cause,  and  in  the  second  ef- 
fect as  in  the  first  effect ;  hence,  if  any  term  is  wanting,  write 
x  in  its  place.     Multiply  together  all  of  the  numbers  within 
the  vertical  lines,  and  also  all  those  without  the  lines  (cancel- 
ing previously,  if  possible],  and  divide  the  product  of  those 
numbers  which  do  not  contain  x  by  the  product  of  the  others 
in  which  x  occurs,  and  the  result  will  be  the  value  of  x. 

334.  EXAMPLE.— If  40  men  can  dig  a  ditch  720  feet  long,  5  feet 
wide  and  4  feet  deep  in  a  certain  time,  how  long  a  ditch  6  feet  deep  and 
3  feet  wide  could  24  men  dig  in  the  same  time  ? 

SOLUTION.— Here  40  men  and  24  men  are  the  causes  and  the  two 
ditches  are  the  effects.     Hence, 


24     = 


2  whence,  jr  = 


=  480  feet.     Ans. 


335.  EXAMPLE. — The  volume  of  a  cylinder  varies  directly  as  its 
length  and  directly  as  the  square  of  its  diameter.  If  the  volume  of  a 
cylinder  10  inches  in  diameter  and  20  inches  long  is  1,570.8  cubic  inches, 
what  is  the  volume  of  another  cylinder  16  inches  in  diameter  and 
24  inches  long  ? 

SOLUTION. — In  this  example,  either  the  dimensions  or  the  volumes 
may  be  considered  the  causes;  say  we  take  the  dimensions  for  the 
Then,  squaring  the  diameters, 


=  1,670.8 


100 


256 


=     1,670.8 


whence,  ^  = 


— =  4,825.4976  cubic  inches.     Ans. 


ARITHMETIC. 


Ill 


336.  EXAMPLE.— If  a  block  of  granite  8  ft.  long,  5  ft.  wide  and  3 
ft.  thick  weighs  7,200  lb.,  what  will  be  the  weight  of  a  block  of  granite 
12  ft.  long,  8  ft.  wide  and  5  ft.  thick  ? 

SOLUTION.— Taking  the  weights  as  the  effects,  we  have 
4 


=  7,200 


x,  or  x  =  4  x  7,200  =  28,800  pounds.     Ans. 


337.  EXAMPLE.— If  12  compositors  in  30  days  of  10  hours  each 
set  up  25  sheets  of  16  pages  each,  32  lines  to  the  page,  in  how  many 
days  8  hours  long  can  18  compositors  set  up,  in  the  same  type,  64  sheets 
of  12  pages  each,  40  lines  to  the  page  ? 

SOLUTION. — Here  compositors,  days,  and  hours  compose  the  causes, 
and  sheets,  pages,  and  lines  the  effects.  Hence, 


J2,  or   x  =  3  X  10  X  2  =  60  days. 


Ans. 


338.  In  examples  stated  like  that  in  Art.  335,  should 
an  inverse  proportion  occur,  write  the  various  numbers  as 
in  the  preceding  examples,  and  then  transpose  those  num- 
bers which  are  said  to  vary  inversely  from  one  side  of  the 
vertical  line  to  the  other  side. 

EXAMPLE. — The  centrifugal  force  of  a  revolving  body  varies  directly 
as  its  weight,  as  the  square  of  its  velocity  and  inversely  as  the  radius  of 
the  circle  described  by  the  center  of  the  body.  If  the  centrifugal  force 
of  a  body  weighing  15  pounds  is  187  pounds  when  the  body  revolves  in 
a  circle  having  a  radius  of  12  inches,  with  a  velocity  of  20  feet  per  sec- 
ond, what  will  be  the  centrifugal  force  of  the  same  body  when  the  radius 
is  increased  to  18  inches  and  the  speed  is  increased  to  24  feet  per  second  ? 

SOLUTION.— Calling  the  centrifugal  force  the  effect,  we  have, 


15 
20' 
12 


15 
24* 

18 


187 


Transposing  12  and  18  (since  the  radii  are  to  vary  inversely)  and  squar- 
ing 20  and  24, 
W 


26 


=  187 


12 


x,  or  x 


•_'.". 


Ana 


112  ARITHMETIC. 

EXAMPLES  FOR   PRACTICE. 

339*     Solve  the  following  by  compound  proportion: 

1.  If  12  men  dig  a  trench  40  rods  long  in  24  days  of  10  hours  each, 
how  many  rods  can  16  men  dig  in  18  days  of  9  hours  each  ? 

Ans.  36  rods. 

2.  If  a  piece  of  iron  7  ft.  long,  4  in.  wide,  and  6  in.  thick  weighs  600 
lb.,  how  much  will  a  piece  of  iron  weigh  that  is  16  ft.  long,  8  in.  wide 
and  4  in.  thick  ?  Ans.  1,828$  lb. 

3.  If  24  men  can  build  a  wall  72  rods  long,  6  feet  wide,  and  5  feet 
high  in  60  days  of  10  hours  each,  how  many  days  will  it  take  32  men  to 
build  a  wall  96  rods  long,  4  feet  wide  and  8  feet  high,  working  8  hours 
a  day  ?  Ans.  80  days. 

4.  The  horsepower  of  an  engine  varies  as  the  mean  effective  pressure, 
as  the  piston  speed  and  as  the  square  of  the  diameter  of  the  cylinder. 
If  an  engine  having  a  cylinder  14  inches  in  diameter  develops  112 
horsepower  when  the  mean  effective  pressure  is  48  pounds  per  square 
inch  and  the  piston  speed  is  500  feet  per  minute,  what  horsepower  will 
another  engine  develop  if  the  cylinder  is  16  inches  in  diameter,  piston 
speed  is  600  feet  per  minute,  and  mean  effective  pressure  is  56  pounds 
per  square  inch  ?  Ans.  204.8  horsepower. 

5.  Referring  to  the  example  in  Art.  335,  what  will  be  the  volume 
of  a  cylinder  20  inches  in  diameter  and  24  inches  long  ' 

Ans.  7,539.84  cubic  inches. 

G.     Knowing  that  the  product  of  3x5x7X9  is  945,  what  is  the 
product  of  6  X  15  X  14  X  36  ?  Ans.  45,360. 

7.  The  speed  in  miles  per  hour  of  a  locomotive  is  directly  pro- 
portional to  the  diameter  of  its  driving  wheels  and  the  number  of 
revolutions  they  make  in  one  minute.     A  locomotive  with  driving 
wheels  06  inches  in  diameter  runs  29.45  miles  in  an  hour  when  the 
wheels  make  150  revolutions  per  minute ;  how  many  miles  will  be  run 
in  one  hour  by  a  locomotive   having  wheels  72  inches  in  diameter 
running  220  revolutions  per  minute  ?  Ans.  47.12  miles  per  hour. 

8.  The  capacity  of  a  cylindrical  boiler  is  proportional  to  its  length 
and  the  square  of  its  diameter.     A  boiler  12  feet  long  and  48  inches  in 
diameter  will  hold  1,128  gallons;  what  is  the  capacity  of  a  boiler  16  feet 
long  and  42  inches  in  diameter  ?  Ans.   1,151.5  gallons. 

9.  The  power  that  may  be  transmitted  by  a  belt  is  proportional 
to  its  width  and  the  diameter  and  number  of  revolutions  made  by  the 
pulley  on  which  it  runs.     If  a  belt  12  inches  wide  will   transmit  10 
horsepower  when  running  over  a  pulley  20  inches  in  diameter  that 
makes   125  revolutions  per  minute,   how   many  horsepower  may  be 
transmitted  by  a  belt  8  inches  wide  when  running  over  a  pulley  30 
inches  in  diameter  that  makes  200  revolutions  per  minute  ? 

Ans.  16  horsepower. 


ARITHMETIC.  113 

10.  The  load  that  a  beam  supported  at  the  two  ends  will  carry  is 
directly  proportional  to  its  width  and  the  square  of  its  depth,  and 
inversely  proportional  to  its  length.  If  an  oak  beam  8  inches  wide,  12 
inches  deep,  and  15  feet  long  will  safely  carry  a  load  of  13,824  pounds, 
what  is  the  safe  load  that  a  beam  10  inches  wide,  16  inches  deep,  and 
20  feet  long  will  support  ?  Ans.  23,040  pounds. 


MENSURATION  AND  USE  OF  LETTERS 
IN  ALGEBRAIC  FORMULAS. 


FORMULAS. 

340.  The  term  formula,  as  used  in  mathematics  and 
in  technical  books,  may  be  defined  as  a  rule  in  ivliicli  symbols 
are  used  instead  of  words  ;  in  fact,  a  formula  may  be  regarded 
as  a  shorthand  method  of  expressing  a  rule.     Any  formula 
can  be  expressed  in  words,  and  when  so  expressed  it  becomes 
a  rule. 

Formulas  are  much  more  convenient  than  rules;  they 
show  at  a  glance  all  the  operations  that  are  to  be  performed; 
they  do  not  require  to  be  read  three  or  four  times,  as  is  the 
case  with  most  rules,  to  enable  one  to  understand  their 
meaning;  they  take  up  much  less  space,  both  in  the  printed 
book  and  in  one's  note-book,  than  rules;  in  short,  whenever 
a  rule  can  be  expressed  as  a  formula,  the  formula  is  to  be 
preferred. 

As  the  term  "quantity  "  is  a  very  convenient  one  to  use, 
we  will  define  it.  In  mathematics,  the  word  quantity  is 
applied  to  anything  that  it  is  desired  to  subject  to  the  ordi- 
nary operations  of  addition,  subtraction,  multiplication,  etc., 
when  we  do  not  wish  to  be  more  specific  and  state  exactly 
what  the  thing  is.  Thus,  we  can  say  "two  or  more  num- 
bers, "or  "two  or  more  quantities;"  the  word  quantity  is 
more  general  in  its  meaning  than  the  word  number. 

341.  The  signs  used  in  formulas  are  the  ordinary  signs 
indicative  of  operations,  and  the  sighs  of  aggregation.     All 
these  signs  are  explained  in  arithmetic,  but  some  of  them 
will  here  be  explained   in  order  to   refresh   the  student's 
memory. 


116  MENSURATION. 

The  signs  indicative  of  operations  are  six  in  number,  viz. : 
+,  -,  X,  -y,  |  ,  V- 

Division  is  indicated  by  the  sign  -^-,  or  by  placing  a  straight 
line  between  the  two  quantities.  Thus,  25  [  17,  25  /  17, 
and  ff  all  indicate  that  25  is  to  be  divided  by  17.  When 
both  quantities  are  placed  on  the  same  horizontal  line,  the 
straight  line  indicates  that  the  quantity  on  the  left  is  to 
be  divided  by  that  on  the  right.  When  one  quantity  is 
below  the  other,  the  straight  line  between  indicates  that 
the  quantity  above  the  line  is  to  be  divided  by  the  one 
below  it. 

The  sign  (|/)  indicates  that  some  root  of  the  quantity  to 
the  right  is  to  be  taken;  it  is  called  the  radical  sign.  To 
indicate  what  root  is  to  be  taken,  a  small  figure,  called  the 
index,  is  placed  within  the  sign,  this  being  always  omitted 
when  the  square  root  is  to  be  indicated.  Thus,  |/25  indi- 
cates that  the  square  root  of  25  is  to  be  taken ;  $  25  indicates 
that  the  cube  root  of  25  is  to  be  taken ;  etc. 

The  signs  of  aggregation  are  four  in  number;  viz.,  , 

(  ),  [  ],  and  {  },  respectively  called  the  vinculum,  the  pa- 
renthesis, the  brackets,  and  the  brace ;  they  are  used 
when  it  is  desired  to  indicate  that  all  the  quantities  in- 
cluded by  them  are  to  be  subjected  to  the  same  operation. 
Thus,  if  we  desire  to  indicate  that  the  sum  of  5  and  8  is  to 
be  multiplied  by  7,  and  we  do  not  wish  to  actually  add 
5  and  8  before  indicating  the  multiplication,  we  may  employ 
any  one  of  the  four  signs  of  aggregation  as  here  shown : 
B~+~8  X  7,  (5  +  8)  X  7,  [5  +  8]  X  7,  {  5  +  8  }  X  7.  The  vin- 
culum is  placed  above  those  quantities  which  are  to  be  treated 
as  one  quantity  and  subjected  to  the  same  operations. 

While  any  one  of  the  four  signs  may  be  used  as  shown 
above,  custom  has  restricted  their  use  somewhat.  The 
vinculum  is  rarely  used  except  in  connection  with  the  radical 
sign.  Thus,  instead  of  writing  f  (5  +  8),  ^[5  +  8],  or 
^{5  +  8}  for  the  cube  root  of  5  plus  8,  all  of  which  would 
be  correct,  the  vinculum  is  nearly  always  used,  |/5  +  8. 

In  cases  where  but  one  sign  of  aggregation  is  needed  (ex- 
cept, of  course,  when  a  root  is  to  be  indicated),  the  parenthesis 


MENSURATION.  117 

is  always  used.  Hence,  (5  -{-  8)  X  7  would  be  the  usual  way 
of  expressing  the  product  of  5  plus  8,  and  7. 

If  two  signs  of  aggregation  are  needed,  the  brackets  and 
parenthesis  are  used,  so  as  to  avoid  having  a  parenthesis 
within  a  parenthesis,  the  brackets  being  placed  outside.  For 
example,  [(20  —  5)  -4-  3]  X  9  means  that  the  difference  be- 
tween 20  and  5  is  to  be  divided  by  3,  and  this  result  multi- 
plied by  9. 

If  three  signs  of  aggregation  are  required,  the  brace, 
brackets,  and  parenthesis  are  used,  the  brace  being  placed 
outside,  the  brackets  next,  and  the  parenthesis  inside.  For 
example,  {[(20  —  5)  -5-  3]  X  9  —  21}  -4-  8  means  that  the 
quotient  obtained  by  dividing  the  difference  between  20 
and  5  by  3  is  to  be  multiplied  by  9,  and  that  after  21  has 
been  subtracted  from  the  product  thus  obtained,  the  result 
is  to  be  divided  by  8. 

Should  it  be  necessary  to  use  all  four  of  the  signs  of  aggre- 
gation, the  brace  would  be  put  outside,  the  brackets  next,  the 
parenthesis  next,  and  the  -vinculum  inside.  For  example, 
{[(20-5^3)  X  9-  21]  -4-8}  X  12. 

As  stated  in  arithmetic,  when  several  quantities  are  con- 
nected by  the  various  signs  indicating  addition,  subtraction, 
multiplication,  and  division,  the  operation  indicated  by  the 
sign  of  multiplication  must  always  be  performed  first.  Thus, 
2  +  3x4  equals  14,  3  being  multiplied  by  4,  before  adding 
to  2.  Similarly,  10  -=-  2  X  5  equals  1,  since  2x5  equals  10, 
and  10  -~  10  equals  1.  Hence,  in  the  above  case,  if  the  brace 
were  omitted,  the  result  would  be  £,  whereas,  by  inserting 
the  brace,  the  result  is  36. 

Following  the  sign  of  multiplication  comes  the  sign  of 
division  in  order  of  importance.  For  example,  5  —  9  -h  3 
equals  2,  9  being  divided  by  3  before  subtracting  from  5. 
The  signs  of  addition  and  subtraction  are  of  equal  value; 
that  is,  if  several  quantities  are  connected  by  plus  and  minus 
signs,  the  indicated  operations  may  be  performed  in  the 
order  in  which  the  quantities  are  placed. 

There  is  one  other  sign  used,  which  is  neither  a  sign  of 
aggregation  nor  a  sign  indicative  of  an  operation  to  be 


118  MENSURATION. 

performed;  it  is  (  =  ),  and  is  called  the  sign  of  equality  ;  it 
means  that  all  on  one  side  of  it  is  exactly  equal  to  all  on  the 
other  side.  For  example,  2=2,  5  —  3  =  2,  5x(14— 9)  =  25. 

342.  Having  called  particular  attention  to  certain  signs 
used  in  formulas,  the  formulas  themselves  will  now  be  ex- 
plained. First,  consider  the  well-known  rule  for  finding  the 
horsepower  of  a  steam-engine,  which  may  be  stated  as  follows : 

Divide  the  continued  product  of  the  mean  effective  pressure 
in  pounds  per  square  inch,  the  lengtJi  of  the  stroke  in  feet,  the 
area  of  the  piston  in  square  inches,  and  the  number  of  strokes 
per  minute,  by  33,000  ;  the  result  will  be  the  horsepower. 

This  is  a  very  simple  rule,  and  very  little,  if  anything, 
will  be  saved  by  expressing  it  as  a  formula,  so  far  as  clear- 
ness is  concerned.  The  formula,  however,  will  occupy  a 
great  deal  less  space,  as  we  shall  show. 

An  examination  of  the  rule  will  show  that  four  quantities 
(viz.,  the  mean  effective  pressure,  the  length  of  the  stroke, 
the  area  of  the  piston,  and  the  number  of  strokes)  are  mul- 
tiplied together,  and  the  result  is  divided  by  33,000.  Hence, 
the  rule  might  be  expressed  as  follows: 

mean  effective  pressure          stroke 

Horsepower  =  ..  .     .  .  X  ,-     r     .s 

(in  pounds  per  square  inch)      (in  feet) 

area  of  piston         number  of  strokes 
(in  square  inches)  (per  minute) 

This  expression  could  be  shortened  by  representing  each 
quantity  by  a  single  letter;  thus,  representing  horsepower 
by  the  letter  "//,"  the  mean  effective  pressure  in  pounds 
per  square  inch  by  "/*,"  the  length  of  stroke  in  feet  by  "  L," 
the  area  of  the  piston  in  square  inches  by  "A,"  the  number 
of  strokes  per  minute  by  "YV,"and  substituting  these  letters 
for  the  quantities  that  they  represent,  the  above  expression 
would  reduce  to 

rr=-pX  L  X  A  xN 
33,000 

a  much  simpler  and  shorter  expression.  This  last  expres- 
sion is  called  a  formula. 


X 


MENSURATION.  118* 

The  formula  just  given  shows,  as  we  stated  in  the  begin- 
ning, that  a  formula  is  really  a  shorthand  method  of  ex- 
pressing a  rule.  It  is  customary,  however,  to  omit  the  sign 
of  multiplication  between  two  or  more  quantities  when  they 
are  to  be  multiplied  together,  or  between  a  number  and  a 
letter  representing  a  quantity,  it  being  always  understood 
that,  when  two  letters  are  adjacent  with  no  sign  between 
them,  the  quantities  represented  by  these  letters  are  to  be 
multiplied.  Bearing  this  fact  in  mind,  the  formula  just 
given  can  be  further  simplified  to 

PLAN 
33,000  ' 

The  sign  of  multiplication,  evidently,  can  not  be  omitted 
between  two  or  more  numbers,  as  it  would  then  be  impos- 
sible to  distinguish  the  numbers.  A  near  approach  to  this, 
however,  may  be  attained  by  placing  a  dot  between  the 
numbers  which  are  to  be  multiplied  together,  and  this  is 
frequently  done  in  works  on  mathematics  when  it  is  desired 
to  economize  space.  In  such  cases  it  is  usual  to  put  the  dot 
higher  than  the  position  occupied  by  the  decimal  point. 
Thus,  2- 3  means  the  same  as  2  X  3;  542-749-1,006  indicates 
that  the  numbers  542,  749,  and  1,006  are  to  be  multiplied 
together. 

It  is  also  customary  to  omit  the  sign  of  multiplication  in 
expressions  similar  to  the  following:  a  X  ^b  +  c,  3  X  (b  +  0» 
(b  +  c)  X  ft,  etc.,  writing  them  a  \/b  +  ct  3  (b  +  c),  (b  +  c)  a, 
etc.  The  sign  is  not  omitted  when  several  quantities  are 
included  by  a  vinculum,  and  it  is  desired  to  indicate  that 
the  quantities  so  included  are  to  be  multiplied  by  another 
quantity.  For  example,  3X^  +  ^",  b  -j-  c  X  a,  tfb  +  c  X  a, 
etc.,  are  always  written  as  here  printed. 

343.  Before  proceeding  further,  we  will  explain  one 
other  device  that  is  used  by  formula  makers  and  which  is 
apt  to  puzzle  one  who  encounters  it  for  the  first  time — it  is 
the  use  of  what  mathematicians  call  primes  and  subs.,  and 
what  printers  call  superior  and  inferior  characters.  As  a 
rule,  formula  makers  designate  quantities  by  the  initial 


118*  MENSURATION. 

letters  of  the  names  of  the  quantities.  For  example,  they 
represent  volume  by  v,  pressure  by  /,  height  by  At  etc. 
This  practice  is  to  be  commended,  as  the  letter  itself  serves 
in  many  cases  to  identify  the  quantity  which  it  represents. 
Some  authors  carry  the  practice  a  little  further,  and  repre- 
sent all  quantities  of  the  same  nature  by  the  same  letter 
throughout  the  book,  always  having  the  same  letter  repre- 
sent the  same  thing.  Now,  this  practice  necessitates  the 
use  of  the  primes  and  subs,  above  mentioned,  when  two 
quantities  have  the  same  name  but  represent  different  things. 
Thus,  consider  the  word  pressure  as  applied  to  steam,  at  dif- 
ferent stages  between  the  boiler  and  the  condenser.  First, 
there  is  absolute  pressure,  which  is  equal  to  the  gauge  pres- 
sure in  pounds  per  square  inch  plus  the  pressure  indicated 
by  the  barometer  reading  (usually  assumed  in  practice  to  be 
14.7  pounds  per  square  inch,  when  a  barometer  is  not  at 
hand).  If  this  be  represented  by/,  how  shall  we  represent 
the  gauge  pressure  ?  Since  the  absolute  pressure  is  always 
greater  than  the  gauge  pressure,  suppose  we  decide  to  repre- 
sent it  by  a  capital  letter,  and  the  gauge  pressure  by  a  small 
(lower-case)  letter.  Doing  so,  P  represents  absolute  pres- 
sure, and  /,  gauge  pressure.  Further,  there  is  usually  a 
"  drop "  in  pressure  between  the  boiler  and  the  engine,  so 
that  the  initial  pressure,  or  pressure  at  the  beginning  of  the 
stroke,  is  less  than  the  pressure  at  the  boiler.  How  shall 
we  represent  the  initial  pressure  ?  We  may  do  this  in  one 
of  three  ways  and  still  retain  the  letter/  or  P  to  represent 
the  word  pressure:  First,  by  the  use  of  the  prime  mark; 
thus,  p'  or  P  (fta&p  prime  and  P  major  prime)  may  be  con- 
sidered to  represent  the  initial  gauge  pressure,  or  the  initial 
absolute  pressure.  Second,  by  the  use  of  sub.  figures;  thus, 
/>,  or  Pt  (read  p  sub.  one  and  P  major  sub.  one}.  Third,  by 
the  use  of  sub.  letters ;  thus,  pt  or  Pt  (read  p  sub.  i  and  P 
major  sub.  i).  In  the  same  manner/"  (read  /  second),  pv  or 
pr  might  be  used  to  represent  the  gauge  pressure  at  release, 
etc.  The  sub.  letters  have  the  advantage  of  still  further 
identifying  the  quantity  represented;  in  many  instances, 
however,  it  is  not  convenient  to  use  them,  in  which  case 


MENSURATION.  lift: 

primes  and  subs,  are  used  instead.  The  prime  notation 
may  be  continued  as  follows:  /'",  /'",  /",  etc.  ;  it  is  inad- 
visable to  use  superior  figures,  for  example,  /',  /",  /*,  pat 
etc.,  as  they  are  liable  to  be  mistaken  for  exponents. 

The  main  thing  to  be  remembered  by  the  student  is  that 
when  a  formula  is  given  in  which  the  same  letters  occur  sev- 
eral times,  all  like  letters  having  the  same  primes  or  subs. 
represent  the  same  quantities,  '<.vhile  those  which  differ  in  any 
respect  represent  different  quantities.  Thus,  in  the  formula 


wlt  w»  and  «%  represent  the  weights  of  three  different 
bodies;  st,  stt  and  sa,  their  specific  heats;  and  t^  /2,  and  /8, 
their  temperatures;  while  t  represents  the  final  temperature 
after  the  bodies  have  been  mixed  together.  It  should  be 
noted  that  those  letters  having  the  same  subs,  refer  to  the 
same  bodies.  Thus,  w^  slt  and  /,  all  refer  to  one  of  the 
three  bodies  ;  zv^,  s^  t^  to  another  body  ;  etc. 

It  is  very  easy  to  apply  the  above  formula  when  the 
values  of  the  quantities  represented  by  the  different  letters 
are  known.  All  that  is  required  is  to  substitute  the  numeri- 
cal values  of  the  letters,  and  then  perform  the  indicated 
operations.  Thus,  suppose  that  the  values  of  w,,  j,,  and  tl 
are,  respectively,  2  pounds,  .0951,  and  80°;  of  wa,  sv  and  /a, 
7.8  pounds,  1,  and  80°;  and  of  w3,  jjt  and  /3,  3£  pounds, 
.1138,  and  780°;  then,  the  final  temperature  /is,  substituting 
these  values  for  their  respective  letters  in  the  formula, 

2X  .0951  X  80+  7.8  X  1  X  80  +  3J  X  .1138  X  780  _ 

2  X  .0951  +  7.8  X  1  +  3±  X  .1138 
15.216  +  624  +  288.483  _  927.  699  0 

.1902  +  7.  8  +.36985     "8.36005 

In  substituting  the  numerical  values,  the  signs  of  multi- 
plication are,  of  course,  written  in  their  proper  places;  all 
the  multiplications  are  performed  before  adding,  accord- 
ing  to  the  rule  previously  given. 


118//  MENSURATION. 

344.  The  student  should  now  be  able  to  apply  any 
formula  involving  only  algebraic  expressions  that  he  may 
meet  with,  and  which  does  not  require  the  use  of  logarithms 
for  its  solution.  We  will,  however,  call  his  attention  to 
one  or  two  other  facts  that  he  may  have  forgotten. 

Expressions  similar  to  —  —  sometimes  occur,  the  heavy  line 

"25* 

indicating  that  160  is  to  be  divided  by  the  quotient  obtained 
by  dividing  660  by  25.  If  both  lines  were  light  it  would 
be  impossible  to  tell  whether  160  was  to  be  divided  by 

/»/>/-v  -•   />/% 

—  -  ,  or  whether  -—  -  was  to  be  divided  by  25.  If  this  latter 
25  ooO 

160 

~{*f*(\ 

result  were  desired,  the  expression  would  be  written  -—  .     In 

ywO 

every  case  the  heavy  line  indicates  that  all  above  it  is  to  be 
divided  by  all  below  it. 

In   an  expression  like   the  following,  -  ;-—  -,  the  heavy 

7  +  ^H 
r  25 

line  is  not  necessary,  since  it  is  impossible  to  mistake 
the  operation  that  is  required  to  be  performed.  But,  since 


,  660      175  +  660    .,  175  +  660,         ,   660 

7  +  ^5-=         25        '  lf  WC   substitute        ^        for  7+  —  , 

the  heavy  line  becomes  necessary  in  order  to   make   the 
resulting  expression  clear.     Thus, 

160  160        _  160 

660  ~  175  +  660  ~  83F 
+  25  25  25 

Fractional  exponents  are  sometimes  used  instead  of  the 
radical  sign.  That  is,  instead  of  indicating  the  square, 
cube,  fourth  root,  etc.,  of  some  quantity,  as  37,  by  4/37, 
4^37,  4/37,  etc.,  these  roots  are  indicated  by  37*,  37^,  37*, 
etc.  Should  the  numerator  of  the  fractional  exponent  be 
some  quantity  other  than  1,  this  quantity,  whatever  it  may 
be,  indicates  that  the  quantity  affected  by  the  exponent  is 
to  be  raised  to  the  power  indicated  by  the  numerator;  the 


MENSURATION.  U8e 

denominator  is  always  the  index  of  the  root.  Hence, 
instead  of  writing  1/37*  for  the  cube  root  of  the  square  of  37, 
it  may  be  written  37*,  the  denominator  being  the  index  of 
the  root;  in  other  words,  1/37*  =  37*.  Likewise,  {/(I  +  a*  b)* 
may  also  be  written  (1  -f  «2  £)*,  a  much  simpler  expression. 

345.  We  will  now  give  several  examples  showing  how 
to  apply  some  of  the  more  difficult  formulas  that  the  student 
may  encounter. 

1.  The  area  of  any  segment  of  a  circle  that  is  less  than 
(or  equal  to)  a  semicircle  is  expressed  by  the  formula 


in  which  A  =  area  of  segment; 

*  =3.1416; 

r  =  radius; 

E  —  angle  obtained  by  drawing  lines  from  the  cen- 
ter to  the  extremities  of  arc  of  segment; 

c  =  chord  of  segment; 
and  h  =  height  of  segment. 

EXAMPLE.  —  What  is  the  area  of  a  segment  whose  chord  is  10  inches 
long,  angle  subtended  by  chord  is  83.46°,  radius  is  7.5  inches,  and 
height  of  segment  is  1.91  inches  ? 

SOLUTION.  —  Applying  the  formula  just  given, 

*r*E       c  3.  1416X7.  5*  X  83.  46      10 


2  360 

40.  968  —  27.  95  =  13.  018  square  inches,  nearly.     Ans. 

2.  The  area  of  any  triangle  may  be  found  by  means  of 
the  following  formula,  in  which  A  =  the  area,  and  a,  by  and  c 
represent  the  lengths  of  the  sides: 


EXAMPLE. — What  is  the  area  of  a  triangle  whose  sides  are  21  feet, 
46  feet,  and  50  feet  long  ? 

SOLUTION. — In  order  to  apply  the  formula,  suppose  we  let  a  repre- 
sent the  side  that  is  21  feet  long  ;  />,  the  side  that  is  50  feet  long  ;  and 
c,  the  side  that  is  46  feet  long.  Then  substituting  in  the  formula, 


118/  MENSURATION. 


=  25^/441  —  8.252  =  25  |/441  —  68.0625  =  25  4/372.9375 
=  25  X  19.312  =  482.8  square  feet,  nearly.     Ans. 
The  operations  in  the  above  examples  have  been  extended 
much  farther  than  was  necessary ;  it  was  done  in  order  to 
show  the  student  every  step  of  the  process.     The  last  for- 
mula is  perfectly  general,  and  the  same  answer  would  have 
been  obtained  had  the  50-foot  side  been  represented  by  «, 
the  46-foot  side  by  $,  and  the  21-foot  side  by  c. 

3.     The    Rankine-Gordon  formula  for   determining    the 
least  load  in  pounds  that  will  cau^c  a  long  column  to  break  is 

p__SA_ 

•*    —~  ~~  rr~  * 


in  which  P  =  load  (pressure)  in  pounds; 

5"  =  ultimate  strength  (in  pounds  per  square  inch) 

of  the  material  composing  the  column; 
A  =  area    of    cross-section   of    column    in    square 

inches  ; 

q  =  a  factor  (multiplier)  whose  value  depends  upon 
the  shape  of  the  ends  of  the  column  and  on 
the  material  composing  the  column  ; 
/  =  length  of  column  in  inches; 

and  G  =  least   radius  of  gyration   of   cross-section  of 

column. 

The  values  of  S,  q,  and  G*  are  given  in  printed  tables  in 
books  in  which  this  formula  occurs. 

EXAMPLE.—  What  is  the  least  load  that  will  break  a  hollow  wrought- 
iron  column  whose  outside  diameter  is  14  inches;  inside  diameter, 
11  inches;  length,  20  feet,  and  whose  ends  are  flat  ? 

SOLUTION.—  For   steel,    5  =  150,000,    and    q  =  -HTTT^  for  flat-ended 


steel  columns  ;  A,  the  area  of  the  cross-section,  =  .7854  (d^  —  </a8)  = 
.7854  (14*  —  11*),  </!  and  d*  being   the   outside  and   inside  diameters, 


MENSURATION. 

respectively ;  /  =  20  X  12  =  240  inches ;  and  G*  =  **** .+.  ^**  =  14"  +  11* 

lo  16 

Substituting  these  values  in  the  formula, 

SA       _  150,000  X  .7854  (14* -11*) 

'         240* 

25,000       14*  +  11- 

150,000  X  58.905      8,835,750 

1  +  .1163          =  ^163- =  7,915,211  pounds.     Ans. 

4.      EXAMPLE. — When   A  =  10,  2?  =  8,  C=5,  and    D  —  4,    what    is 
the  value  of  E  in  the  following  : 


SOLUTION. — (a)  Substituting, 


To  simplify  the  denominator,  square  the  4  and  5,  add  the  resulting 
fraction  to  2,  and  multiply  by  10.     Simplifying,  we  have 


2  + 


/    160  /160  _      /2QO 

™\~\/  wx-~i^  —  ~V    33' 

25j  X  25  25 


Reducing  the  fraction  to  a  decimal  before  extracting  the  cube  root. 

£=   1/6.0606  =  1.823.     Ans. 
(b)  Substituting, 


7  +  17.066+  =  24.066+ 
10-1/4  8 

345 j. — In  the  preceding  pages,  the  unknown  quantity 
has  always  been  represented  by  the  single  letter  at  the  left 
of  the  sign  of  equality,  while  the  letters  at  the  right  have 
represented  known  values  from  which  the  required  values 
could  be  found.  It  is  possible,  however,  to  find  the  value  of 
the  quantity  represented  by  any  letter  in  a  formula,  if  the 
values  represented  by  all  the  others  are  known.  For  example, 


118/*  MENSURATION. 

let  it  be  required  to  find  how  many  strokes  per  minute  an  en- 
gine having  a  piston  area  of  78.54  sq.  in.  must  make  in  order 
to  develop  60  horsepower,  if  the  mean  effective  pressure  is 
40  Ib.  per  sq.  in.,  and  the  length  of  stroke  1£  ft.  By  substi- 

„     PLAN 

tutmg  the  given  values  in  the  formula  H  —  ,  we  have 

00,000 

40  X  1JX  78.54  X^V 

33,000 
in  which  N,  the  number  of  strokes,  is  to  be  found. 

But  it  is  evident  that  the  expression  on  the  right  of  the 

,  ,        40  xliX  78.54       ,r       . 

sign  of  equality  is  equal  to  --  —         --  X  -/V,  a  fraction 

oo,  000 

whose  numerator  is  composed  of  3  factors.  Reducing  the 
numerator  to  a  single  number  by  performing  the  indicated 
multiplications,  we  obtain,  after  canceling, 


If  60  equals  .119  N,  then  N  equals  60  divided  by  .  119  ;  hence, 

60 
N=  —  —  =  504.2  revolutions  per  minute. 

.  J.  J.  t) 

The  method  of  procedure  is  essentially  the  same  when  the 
unknown  quantity  occurs  in  the  denominator  of  a  formula. 

Thus,  in  the  formula/"  =  --  -,  suppose  that/"—  375,  m  =  1.25, 

and  v  =  60.     Then,  substituting, 

_  1.25  X  60"  _  4,500 

o  i  O  —  —  'o 

T  T 

But  if  375  equals  4,500  divided  by  r,  then  375  X  r  =  4,500; 
hence,  r  must  equal  4,500  divided  by  375,  or  r  =    '        =  12. 

O  4  0 

EXAMPLES   FOR   PRACTICE. 

Find  the  numerical  values  of  x  in  the  following  formulas,   when 
A  =  9,  J3  =  8,  d=  10,  e  =  3,  and  c  =  2  : 

!•     *=  Ans'   *  = 


2.     x  =  Ans. 


MENSURATION. 


119 


Ans.  x  —  29. 

Ans.  x=2. 

Ans.  .r  =  12^. 

Ans.  .r=  .396+. 


MENSURATION. 

346.     Mensuration  treats  of  the  measurement  of  lines, 
angles,  surfaces,  and  solids. 


FIG.  2. 


LINES    AND    ANGLES. 

347.  A  straight  line  is  one   that  does  not  change  its 
direction  throughout  its  whole  length.     To  distinguish  one 
straight  line  from  another,  its  two  ex- 
treme points  are  designated  by  letters.    A —  — B 
The  line  shown  in  Fig.  1  would  be  called                 FIG.  i. 

the  line  A  B. 

348.  A  curved  line   changes  its 
direction  at  every  point.      Curved  lines 
are  designated  by  three  or  more  letters, 
as  the  curved  line  ABC,  Fig.  2. 

349.  Parallel  lines  (Fig.  3)  are 
those  which   are  equally  distant  from 
each  other  at  all  points. 

350.  A  line  is  perpendicular  to 
another    (see   Fig.   4)   when   it    meets 
that  line  so  as  not  to  incline  towards  it 
on  either  side. 

351.  A  vertical  line  is  one  that 
points  towards  the  center  of  the  earth, 
and  is  also  known  as  a.  plumb  line. 

352.  A  horizontal  line  (see  Fig. 
5)  is  one  that  makes  a  right  angle  with 

any  vertical  line.  FIG.  6. 


120 


MENSURATION. 


353.  An  angle  is  the  opening  between  two  lines  which 
intersect  or  meet ;  the  point  of  meeting  is  called  the  vertex 

of  the  angle.  Angles  are  distinguished 
by  naming  the  vertex  and  a  point  on  each 
line.  Thus,  in  Fig.  6,  the  angle  formed 
by  the  lines  A  B  and  C  B  is  called  the 
angle  A  B  C,  or  the  angle  C  B  A  ;  the 

letter  at  the  vertex  is  always  placed  in  the  middle.  When 
an  angle  stands  alone  so  that  it  can  not  be  mistaken  for  any 
other  angle,  only  the  vertex  letter  need  be  used.  Thus,  the 
angle  referred  to  might  be  designated  simply  as  the  angled. 

354.  If   one   straight   line  meets 
another  straight  line  at  a  point  between 
its  ends,  as  in  Fig.  7,  two  angles,  ABC 
and  A  B  D,  are  formed,  which  are  called 
adjacent  angles. 


B 

FIG.  7. 


355.     When  these  adjacent  angles, 
A  B  Cand  A  B  D,  are  equal,  as  in  Fig. 
_D  8,  they  are  called  right  angles. 


FIG.  8. 


356.  An  acute  angle  is  less  than 
a  right  angle.  ABC,  Fig.  9,  is  an  acute 
angle. 


FIG.  9. 


357.  An  obtuse  angle  is  greater 
than  a  right  angle.  A  B  D  (Fig.  10) 
is  an  obtuse  angle. 


358.     A  circle    (see  Fig.   11)  is  a  figure  f  \ 

bounded  by  a  curved  line,  called  the  circum-  /  .1 

ference,  every  point  of  which  is  equally  dis-  V  I 

tant  from  a  point  within,  called  the  center.     \^^^./ 

FIG.  11. 


MENSURATION.  121 

359.  An  arc  of  a  circle  is  any  part  of  &.^ 

its  circumference;  thus  a  e  b,  Fig.  12,  is  an      /x          ^s* 
arc  of  the  circle.  /  \ 

The  circumference  of  every  circle  is  con-    \ 
sidered  to  be  divided  into  3GO  equal  parts,  or    \  /' 

arcs,   called  degrees ;  every  degree  is  sub- 
divided into  60  equal  parts,  called  minutes, 
and  every  minute  is  again  divided  into  60  equal  parts,  called 
seconds. 

Since  1  degree  is  — —  of  any  circumference,  it  follows  that 
ooO 

the  length  of  a  degree  will  be  different  in  circles  of  different 
sizes,  but  the  proportion  of  the  length  of  an  arc  of  one 
degree  to  the  whole  circumference  will  always  be  the  same, 

viz.,  - — of  the  circumference. 
o60 

Degrees,  minutes,  and  seconds  are  denoted  by  the  symbols 
°,  ',  ".  Thus,  the  expression  37°  14'  44"  is  read  37  degrees, 
14  minutes,  and  44  seconds. 

360.  The  arcs  of  circles  are  used  to  measure  angles. 

An  angle  having  its  vertex  at  the  cen- 
ter of  a  circle  is  measured  by  the  arc 
included  between  its  sides;  thus,  in 
Fig.  13,  the  arc  F  B  measures  the  angle 
FOB.  If  the  arc  F  B  contains  20°, 

20 

or  77—  of  the  circumference,  the  angle 
o60 

FOB  would  be  an  angle  of  20° ;  if  it 
FIG.  13.  contained  20°  14'  18",  it  would  be  an 

angle  of  20°  14'  18",  etc. 

In  the  figure,  if  the  line  C  D  be  drawn  perpendicular  to 
A  B,  the  adjacent  angles  will  be  equal,  and  the  circle  will 
be  divided  into  four  equal  angles,  each  of  which  will  be  a 

360° 
right  angle.     A  right  angle,  therefore,  is  an  angle  of  — — , 

or  90° ;  two  right  angles  are  measured  by  180°,  or  half  the 
circumference,  and  four  right  angles  by  the  whole  circum- 
ference, or  360°.  One-half  of  a  right  angle,  as  £  O  B,  is  an 


122  MENSURATION. 

angle  of  45°.  An  acute  angle  may  now  be  defined  as  an 
angle  of  less  than  90°,  and  an  obtuse  angle  as  one  of  more 
than  90°.  These  values  are  important,  and  should  be 
remembered. 

361 .  From  the  foregoing  it  will  be  evident  that  if  a  num- 

ber of  straight  lines  on  the  same  side 
of  a  given  straight  line  meet  at  the 
same  point,  the  sum  of  all  the  angles 
formed  is  equal  to  two  right  angles,  or 
180°.  Thus,  in  Fig.  14,  angles  COB 
+DOC+EOD+FOE+AOF 
=  2  right  angles,  or  180°. 

362.  Also,  if  through  a  given  point 
any  number  of  straight  lines  be  drawn, 
the  sum  of  all  the  angles  formed  about 
the   points  of  intersection   equals  four 
right  angles,  or  360°.     Thus,  in  Fig.  15, 
a.nglesffOF+FOC+COA  +  AOG 
+  GO  E+EO  D  +  DO  B  +  BO  H 
=  four  right  angles,  or  360°. 

EXAMPLE. — In  a  fly-wheel,  with  12  arms,  how  many  degrees  are  there 
in  the  angle  included  between  the  center  lines  of  any  two  arms,  the 
arms  being  spaced  equally  ? 

SOLUTION. — Since  there  are  12  arms,  there  are  12  angles,  which 

Qfift0 

together  equal  360°.     Hence,  one  ahgle  equals  TV  of  360°,  or— ;>-:=  30°. 

Ans. 

EXAMPLES    FOR     PRACTICE. 

1.  How  many  seconds  in  32°  14'  6*  ?  Ans.  116,046  sec. 

2.  How  many  degrees,    minutes,  and  seconds  do  38,582   seconds 
amount  to  ?  Ans.  10°  43'  2". 

3.  How  many  right  angles  are  there  in  an  angle  of  170°  ? 

Ans.  If  right  angles. 

4.  In  a  pulley  with  five  arms,  what  part  of  a  right  angle  is  included 
between  the  center  lines  of  any  two  arms  ?         Ans.  £  of  a  right  angle. 

5.  If  one  straight  line  meets  another  so  as  to  form  an  angle  of  20° 
10',  what  does  its  adjacent  angle  equal  ?  Ans.  159°  50'. 

6.  If  a  number  of  straight  lines  meet  a  given  straight  line  at  a 
given  point,  all  being  on  the  same  side  of  the  given  line,  so  as  to  form 
six  equal  angles,  how  many  degrees  are  there  in  each  angle  ?       Ans.  30°. 


MENSURATION. 


123 


QUADRILATERALS. 

363.  A  plane   figure  is  any  part  of  a  plane  or  flat 
surface  bounded  by  straight  or  curved  lines. 

364.  A  quadrilateral   is  a  plane  figure  bounded  by 
four  straight  lines. 

365.  A  parallelogram   is  a  quadrilateral   whose  op- 
posite sides  are  parallel. 

There  are  four  kinds  of  parallelograms:  the  square,  the 
rectangle,  the  rhombus,  and  the  rhomboid. 


366.     A  rectangle  (Fig.  16)  is  a  parallelo- 
gram whose  angles  are  all  right  angles. 


367.     A  square  (Fig.  17)  is  a  rectangle 
whose  sides  are  all  of  the  same  length. 


FIG.  IT. 

368.  A  rhomboid  (Fig.  18)  is 
a  parallelogram  whose  opposite 
sides  are  equal  and  parallel,  and 
whose  angles  are  not  right  angles. 


A 


369.  A  rhombus  (Fig.  19) 
is  a  parallelogram  having  equal 
sides,  and  whose  angles  are  not 
right  angles. 


FIG 


370.  A  trapezoid  (Fig.  20)  is 

a  quadrilateral  which  has  only  two     ^ 
of  its  sides  parallel. 

371.  The  altitude  of  a  parallelogram  is  the  perpen- 
dicular distance  between  the  parallel  sides,  as  shown  by  the 
dotted  lines  in  Figs.  18,  19,  and  20. 

372.  The  base  of  any  plane  figure  is  the  side  on  which 
it  is  supposed  to  stand. 


124  MENSURATION. 

373.  The  area  of  a  surface  is  expressed  by  the  number 
of  unit  squares  it  will  contain. 

374.  A  unit  square  is  the  square  'having  a  unit  for 
its  side.     For  example,  if  the  unit  is  1  inch,  the  unit  square 
is  the  square  each  of  whose  sides  measures  1  inch  in  length, 
and  the  area  of  a  surface  would  be  expressed  by  the  number 
of  square  inches  it  would  contain.     If  the  unit  were  1  foot, 
the  unit  square  would  measure  1  foot  on  each  side,  and  the 
area  of  the  given  surface  would  be  the  number  of  square 
feet  it  would  contain,  etc. 

The  square  that  measures  one  inch  on  a  side  is  called  a 
square  inch,  and  the  one  that  measures  one  foot  on  a  side 
is  called  a  square  foot.  Square  inch  and  square  foot  are 
abbreviated  to  sq.  in.  and  sq.  ft. 

375.  To  find  the  area  of  any  parallelogram: 
Rule  44. — Multiply  the  base  by  the  altitude. 

NOTE. — Before  multiplying,  the  base  and  altitude  must  be  reduced 
to  the  same  kind  of  units;  that  is,  if  the  base  should  be  given  in  feet 
and  the  altitude  in  inches,  they  could  not  be  multiplied  together  until 
either  the  altitude  had  been  reduced  to  feet,  or  the  base  to  inches. 
This  principle  holds  throughout  the  subject  of  mensuration. 

EXAMPLE.— The  sides  of  a  square  piece  of  sheet  iron  are  each  10± 
inches  long.  How  many  square  inches  does  it  contain  ? 

SOLUTION.— 10£  inches  =10. 25  inches  when  reduced  to  a  decimal. 
The  base  and  altitude  are  each  10.25  inches.  Multiplying  them 
together,  10.25  X  10.25  =  105.06+  sq.  in.  Ans. 

EXAMPLE. — What  is  the  area  in  square  rods  of  a  piece  of  land  in  the 
shape  of  a  rhomboid,  one  side  of  which  is  8  rods  long,  and  whose 
length,  measured  on  a  line  perpendicular  to  this  side,  is  200  feet  ? 

SOLUTION. — The  base  is  8  rods  and  the  altitude  200  feet.  As  the 
answer  is  to  be  in  rods,  the  200  feet  should  be  reduced  to  rods.  Redu- 
cing 200  -i-  16i  =  200  -r-  ^  =  12.12  rods.  Hence,  area  =  8  X  12.12  =  96.96 
sq.  rd.  Ans. 

376.  To  find  the  area  of  a  trapezoid : 

Rule  45. — Multiply  one -half  the  sum  of  the  parallel  sides 
by  the  altitude. 

EXAMPLE. — A  board  14  feet  long  is  20  inches  wide  at  one  end  and  16 
inches  wide  at  the  other.  If  the  ends  are  parallel,  how  many  square 
feet  does  the  board  contain  ? 


MENSURATION.  125 

SOLUTION.— One-half  the  sum  of  the  parallel  sides  = — ^ —  =  13 

inches  =  l£  feet.     The  length  of  the  board  corresponds  to  the  altitude 
of  a  trapezoid.     Hence,  14  x  H  =  21  sq.  ft.     Ans. 

377.  Having  given  the  area  of  a  parallelogram  and  one 
dimension,  to  find  the  other  dimension : 

Rule  46. — Divide  the  area  by  the  given  dimension. 

EXAMPLE.— What  is  the  width  of  a  parallelogram  whose  area  is  212 
square  feet  and  whose  length  is  26i  feet  ? 

SOLUTION.— 212  -=-  26*  =  212  H-  ^  =  8  feet.     Ans. 

The  following  examples  illustrate  a  few  special  cases: 

EXAMPLE.— An  engine  room  is  22  feet  by  32  feet.     The  engine-bed 
occupies  a  space  of  3  feet  by  12  feet ;  the  fly-wheel  pit,  a  space  of  2  feet 
by  6  feet,  and  the  outer  bearing,  a  space  of  2  feet  by  4  feet.     How 
many  square  feet  of  flooring  will  be  required  for  the  room  ? 
SOLUTION.— Area  of  engine-bed       =  3  X  12=  36  sq.  ft. 
Area  of  fly-wheel  pit    =  2  X    6  =  12  sq.  ft. 
Area  of  outer  bearing  =  2  X    4  =    8  sq.  ft. 
Total,     56  sq.  ft. 

Area  of  engine  room  =  22  X  32  =  704  sq.  ft. 
704  —  56  =  648  square  feet  of  flooring  required.     Ans. 

EXAMPLE. — How  many  square  yards  of  plaster  will  it  take  to  cover 
the  sides  and  ceiling  of  a  room  16  X  20  feet  and  11  feet  high,  having 
four  windows,  each  7x4  feet,  and  three  doors,  each  9x4  feet  over  all, 
the  baseboard  coming  6  inches  above  the  floor? 

SOLUTION.— 

Area  of  ceiling       =     16  X  20  =     320  sq.  ft. 

Area  of  end  walls  =  2(16  X  11)  =     352  sq.  ft. 

Area  of  side  walls  =  2(20  X  11)  =     440  sq.  ft. 

Total  area  =  1,112  sq.  ft. 

From  the  above  must  be  deducted : 

Windows  =  4(7  X  4)  =  112  sq.  ft. 
Doors        =  3(9  X  4)  =  108  sq.  ft. 

Baseboard  less  the  width  of  three  doors  =  (72  -  12)  X  p-  =  30  sq.  ft. 
Total  number  of  feet  to  be  deducted  =  112  +  108  +  30  =  250  sq.  ft. 
Hence,    number    of    square   feet  to  be  plastered  =  1,112  —  250  = 
862  sq.  ft.,  or  95 J  sq.  yd.     Ans. 

EXAMPLE. — How  many  acres  are  contained  in  a  rectangular  tract  of 
land  800  rods  long  and  520  rods  wide  ? 

SOLUTION.— 800  X  520  =  416,000  sq.  rd.  Since  there  are  160  square 
rods  in  one  acre,  the  number  of  acres  =  416,000  -4-  160  =  2,600  acres.  Ans. 


12G 


MENSURATION. 


EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  area  in  square  feet  of  a  rhombus  whose  base  is  84 
inches,  and  whose  altitude  is  3  feet  ?  Ans.  21  sq.  ft. 

2.  A  flat  roof,  46  feet  by  80  feet  in  size,  is  covered  by  tin  roofing 
weighing  one-half  pound  per  square  foot ;  what  is  the  total  weight  of 
thereof?  .     Ans.  l,8401b. 

3.  One  side  of  a  room  measures  16  ft.     If  the  floor  contains  240 
square  feet,  what  is  the  length  of  the  other  side  ?  Ans.  15  ft. 

4.  How  many  square  feet  in  a  board  12  feet  long,  18  inches  wide  at 
one  end,  and  12  inches  wide  at  the  other  end  ?  Ans.  15  sq.  ft. 

5.  How  much  would  it  cost  to  lay  a  sidewalk  a  mile  long  and  8  feet 
6  inches  wide,  at  the  rate  of  20  cents  per  square  foot  ?     How  much  at 
the  rate  of  §1.80  per  square  yard  ?  Ans.  $8,976  in  each  case. 

6.  How  many  square  yards  of  plastering  will  be  required  for  the 
ceiling  and  walls  of  a  room  10  X  15  feet  and  9  feet  high;  the  room  con- 
tains one  door  3£  X  7  feet,  three  windows  3£  X  6  feet,  and  a  baseboard 
8  inches  high?  Ans.  53.5  sq.  yd. 

THE    TRIANGLE. 
378.     A  triangle  is  a  plane  figure  having  three  sides. 

379.  An  isosceles  triangle  is 
one  having  two  of  its  sides  equal; 
see  Fig.  21. 

380.  An  equilateral  triangle 

(Fig.  22)  is  one   having   all  of   its 
FIG.  21.     sides  of  the  same  length. 


381.     A  scalene  triangle  (Fig.  23)  is 
one  having  no  two  of  its  sides  equal. 

382.  A  right-angled  triangle  (Fig. 
24)  is  any  triangle  having  one  right  angle. 
The  side  opposite  the  right  angle  is  called 
the  hypotenuse.  FlG.  34. 

In  any  triangle  the  sum  of  the  three  angles  equals  two 
right  angles,  or  180°.  Thus,  in  Fig. 
25,  the  sum  of  the  angles  A,  B,  and 
C  equals  two  right  angles,  or  180°. 
Hence,  if  any  two  angles  of  a  tri- 
O  angle  are  given  and  it  is  required  to 
find  the  third  angle: 


MENSURATION.  12? 

Rule  47. — Add  together  the  two  given  angles,  and  subtract 
their  sum  from  180°;  the  result  will  be  the  third  angle. 

EXAMPLE. — If  two  angles  of  a  triangle  =  48°  16'  and  47°  50'  respec- 
tively, what  does  the  third  angle  equal  ? 

SOLUTION. — First  reduce  48"  16'  and  47°  50'  to  minutes,  for  conve- 
nience in  adding  and  subtracting  the  angles.  48°  =  48  x  60  =  2,880'; 
2,880'  +  16'  =2,896';  hence,  48°  16'  =  2,896'.  In  like  manner,  47°  50'  = 
47  X  60'  =  2,820'  +  50'  =  2,370'.  Adding  the  two  angles  together,  and 
subtracting  from  180°  reduced  to  minutes.  2,896  +  2,870'  =  5,766'; 
180°  =  180  X  60  =  10,800';  10,800  —  5,766  =  5,034'.  Reducing  this  last 

5  034 
number  to  degrees  and  minutes,     '        =  83f  £°  =  83°  54'.     Hence,  the 

third  angle  in  the  triangle  =  83°  54'.     Ans. 

383.  In  any  right-angled  triangle  there  can  be  but  one 
right  angle,  and  since  the  sum  of  all 
the  angles  is  two  right  angles,  it  is 
evident  that  the  sum  of  the  two  acute 
angles  must  equal  one  right  angle,  or 
90°.  Therefore,  if  in  any  right-angled 


triangle  one  acute  angle  is  known,  to    A  — c 

find  the  other  acute  angle :  FlG-  ~°- 

Rule  48. — Subtract  the  known  acute  angle  from  00°;  the 
result  will  be  the  other  acute  angle. 

EXAMPLE. — If  one  acute  angle,  as  A,  of  the  right-angled  triangle 
ABC,  Fig.  26,  equals  30°,  what  does  the  angle  B  equal  ? 
SOLUTION.  —90°  —  30°  =  60°.    Ans. 

384.  If  a  straight  line  be  drawn 
through  two  sides  of  a  triangle,  parallel 
to  the  third  side,  a  second  triangle 
will  be  formed  whose  sides  will  be  pro- 
portional to  the  corresponding  sides  of 
the  first  triangle.  Thus,  in  the  triangle 
A  B  C,  Fig.  27,  if  the  line  D  E  be  drawn 
parallel  to  the  side  B  C,  the  triangle 
FIG.  27.  c  A  D  E  will  be  formed  and  we  shall  have 

(1)  Side  A  D  :  side  D  E  =  side  A  B  :  side  B  C;  and, 

(2)  Side  A  E  :  side  D  E  =  side  A  C  :  side  B  C;  also, 

(3)  Side  A  D  :  side  A  E  =  side  A  B  :  side  A  C. 


128 


MENSURATION. 


EXAMPLE.— In  Fig.  27,  if  A  B  =  24,  B  C  -  18,  and  D  E  =  8,  what 
does  A  D  equal  ? 

SOLUTION. — Writing  these  values  for  the  sides  in  (1), 

A  D  :  8  =  24  :  18;  whence  A  D  =  ***  8  =  10J.     Ans. 

lo 

385.  In  any  right-angled 
triangle,  the  square  described 
on  the  hypotenuse  is  equal 
to  the  sum  of  the  squares 
described  upon  the  other  two 
sides.  If  A  B  C,  Fig.  28,  is 
a  right-angled  triangle,  right- 
angled  at  B,  then  the  square 
described  upon  the  hypote- 
nuse^ C  is  equal  to  the  sum 
of  the  squares  described  upon 
the  sides  A  B  and  B  C. 
FlG- <3g"  Hence,  having  given  the 

two  sides  forming  the  right  angle  in  a  right-angled  triangle, 

to  find  the  hypotenuse: 

Rule  49. — Square  each  of  the  sides  forming  the  right 
angle;  add  the  squares  together,  and  take  the  square  root  of 
the  sum. 

EXAMPLE.— If  A  B  =  3  inches  and  B  C=  4  inches,  what  is  the  length 
of  the  hypotenuse  A  C1 

SOLUTION.— Squaring  each  of  the  given  sides,  3'2  =  9  and  42  =  16. 
Taking  the  square  root  of  the  sum  of  9  and  16,  the  hypotenuse  = 
4/9  +  16  =  |/25  =  5  inches.  Ans. 

If  the  hypotenuse  and  one  side  are  given,  the  other  side 
can  be  found  as  follows: 

Rule  5O. — Subtract  the  square  of  the  given  side  from  the 
square  of  the  hypotenuse,  and  extract  the  square  root  of  the 
remainder. 

EXAMPLE.— The  side  given  is  3  inches,  the  hypotenuse  is  5  inches  ; 
what  is  the  length  of  the  other  side  ? 

SOLUTION.— 32  =  9;  58  =  25.  2,5  -  9  =  16,  and  the  -j/16  =  4  inches. 
Ans. 


MENSURATION. 


129 


150 


EXAMPLE.— If  from  a  church  steeple  which  is  150  feet  high,  a  rope 
is  to  be  attached  to  the  top,  and  to  a  stake  in  the  ground,  which  is  85 
feet  from  the  center  of  the  base  (the  ground  being 
supposed  to  be  level),  what  must  be  the  length  of 
the  rope  ? 

SOLUTION.— In  Fig.  29,  A  B  represents  the 
steeple,  150  feet  high;  C  a  stake  85  feet  from  the 
foot  of  the  steeple,  and  A  C  the  rope.  Here  we 
have  a  right-angled  triangle,  right-angled  at  B,  and 
A  C  is  the  hypotenuse.  The  square  of  A  B  =  1502= 
22,500;  of  C  B,  85*  =  7,225.  22,500  +  7,225  =  29,725 ; 
'29,725  =  172.4  feet,  nearly.  Ans. 

386.    The  altitude 

of  any  triangle  is  aline, 
as  B  D,  drawn  from  the 
vertex  B  of  the  angle 
opposite  the  base  A  C,  . 
perpendicular  to  the 
base,  as  in  Fig.  30,  or 
to  the  base  extended,  as  in  Fig.  31. 

If  in  any  parallelogram  a  straight  line,  called  the 
diagonal,  be  drawn,  connecting  two 
opposite  corners  it  will  divide  the 
parallelogram  into  two  equal  triangles, 
as  A  D  B  and  D  B  C  in  Fig.  32.  The 
FIG.  32.  area  of  each  triangle  will  equal  one-half 

the  area  of  the  parallelogram,  or  one-half  the  product  of  the 
base  and  the  altitude.    Hence,  to  find  the  area  of  any  triangle : 

Rule  5 1 . — Multiply  the  base  by  the  altitude,  and  divide 
the  product  by  2. 

EXAMPLE.— What  is  the  area  in  square  feet  of  a  triangle  whose  base 
is  18  feet,  and  whose  altitude  is  7  feet  9  inches  ? 

SOLUTION.— 7  feet  9  inches  =  7f  feet  =  -^  feet.     18  X  -^  =  139$,  and 
one-half  of  139$  =  69f  square  feet.     Ans. 

'  To  find  the  altitude  or  base  of  a  triangle,  having  given 
the  area  and  the  base  or  altitude: 

Rule  52. — Multiply  the  area  by  2,  and  divide  by  the  given 
dimension. 


130  MENSURATION. 

EXAMPLE. — What  must  be  the  height  of  a  triangular  piece  of  sheet 
metal  to  contain  100  square  inches,  if  the  base  is  10  inches  long  ? 
SOLUTION.  -100  x  2  =  200 ;  200  -=-  10  =  20  inches.     Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  area  of  a  triangle  whose  base  is  18  feet  long,  and 
whose  altitude  is  10  feet  6  inches?  Ans.  94.5  sq.  ft. 

2.  Two  angles  of  a  scalene  triangle  together  equal  100"  4'.    What  is 
the  size  of  the  third  angle  ?  Ans.   79°  56'. 

3.  One  angle  of  a  right-angled  triangle  equals  20°  10'  5".     What  is 
the  size  of  the  other  acute  angle  ?  Ans.   69°  49'  55". 

4.  A  ladder  65  feet  long  reaches  to  the  top  of  a  wall  when  its  foot 
is  25  feet  from  the  wall.     How  high  is  the  wall  ?  Ans.  60  ft. 

5.  Draw  a  triangle,  and  through  two  of  its  sides  draw  a  line  paral- 
lel to  the  base.    Letter  the  different  lines,  and  then,  without  referring 
to  the  text,  write  out  the  proportions  existing  between  the  sides  of  the 
two  triangles. 

6.  A  triangular  piece  of  sheet  metal  weighs  24  pounds.    If  the  base 
of  the  triangle  is  4  feet  and  its  height  6  feet,  how  much  does  the  metal 
weigh  per  square  foot  ?  Ans.  2  Ib. 

7.  The  area  of  a  triangle  is  16  square  inches.     If  the  altitude  is  4 
inches,  what  does  the  base  measure  ?  Ans.  8  in. 

8.  Two  sides  of  a  right-angled  triangle  are  92  feet  and  69  feet  long. 
How  long  is  the  hypotenuse  ?      Ans.  115  ft. 

POLYGONS. 

387.  A  polygon  is  a  plane  figure  bounded  by  straight 
lines.     The  term  is  usually  applied  to  a  figure  having  more 
than  four  sides.     The  bounding  lines  are  called  the  sides, 
and  the  sum  of  the  lengths  of  all  the  sides  is  called  the 
perimeter  of  the  polygon. 

388.  A  regular  polygon  is  one  in  which  all  of  the 
sides  and  all  of  the  angles  are  equal. 

389.  A  polygon  of  five   sides  is  called   a  pentagon; 
one  with  six  sides  a  bexagon;  one  of  seven  sides  a  hep- 


Pentagon.        Hexagon.        Heptagon.        Octagon.          Decagon.        Dodecagon. 
FIG.  13. 

tagon,  etc.      Regular  polygons  having  from  five  to  twelve 
sides  are  shown  in  Fig.  33.     In  any  polygon,  the  sum  of  all 


MENSURATION. 


131 


,  Fig.  34,  equals 


the  interior  angles,  asA+fi+C+D  + 
180°  multiplied  by  a  number  which  is  two 
less  than  the  number  of  sides  in  the  poly- 
gon. Hence,  to  find  the  size  of  any  one  of 
the  interior  angles  of  a  regular  polygon  : 

Rule  53.  —  Multiply  180°  by  the  num- 
ber of  sides  less  two,  and  divide  the  result 
by  the  number  of  sides;  the  quotient  FlG 

will  be  the  number  of  degrees  in  each  interior  angle. 

EXAMPLE.—  If  Fig.  34  is  a  regular  pentagon,  how  many  degrees  are 
there  in  each  interior  angle  ?    • 

SOLUTION.  —  In  a  pentagon  there  are  five  sides;  hence,  3  —  2  =  3  and 
180  X  3  =  540  ;  540  H-  5  =  108°  in  each  angle.     Ans. 

EXAMPLE.  —  It  is  desired  to  make  a  miter-box 
in  which  to  cut  a  strip  of  molding  to  fit  around 
a  column  having  the  shape  of  a  regular  hexa- 
gon. At  what  angle  should  the  saw  run  across 
the  miter-box  ? 

SOLUTION.—  In  Fig.  35,  let  A  />',  />'  C,  CD, 
etc.,  represent  the  pieces  of  molding  as  they 
will  fit  around  the  column.  First  find  the  size 
of  one  of  the  equal  angles  of  the  polygon  by 
the  above  rule.  Number  of  sides  =  6;  6  —  2  = 
4;  hence,  180  X  4  =  720,  and  720  -s-  6  =  120°  in 

each  angle.  Now,  let  M  N  represent  the  miter-box,  and  O  S  the  direc- 
tion in  which  the  saw  should  run;  then,  A  7>  O  is  the  angle  made  by 
the  saw  with  the  side  of  the  miter-box;  but,  as  the  polygon  is  a  regular 
one,  this  angle  is  one-half  the  interior  angle  A  7>  C,  which  we  have 

found  to  be  120°.  Hence,  the  saw  should  run  at  an  angle  of  -£-  =  60° 
with  the  side  of  the  miter-box. 

39O.  The  area  of  any  regular  polygon 
may  be  found  by  drawing  lines  from  the 
center  to  each  angle,  and  computing  the 
area  of  each  triangle  thus  formed.  Hence, 
to  find  the  area  of  any  regular  polygon: 

Rule  54.  —  Multiply  the  length  of  a  side 
by  half  the  distance  from  the  side  to  the  FIG.  se. 

center,  and  that  product  by  the  number  of  sides.  The  last 
product  will  be  the  area  of  the  figure. 


132 


MENSURATION. 


EXAMPLE.— In  Fig.  36  the  side  B  C  of  the  regular  hexagon  is  12 
inches  and  the  distance  A  O  is  10. 4  inches;  required,  the  area  of  the 
polygon. 

SOLUTION.— 10.4  H-  2  =  5.2;  12  X  5.2  X  6  =  374.4  sq.  in.     Ans. 


391.  To  obtain  the  area  of 
any  irregular  polygon,  draw  diag- 
onals dividing  the  polygon  into 
triangles  and  quadrilaterals,  and 
compute  the  areas  of  these  sepa- 
rately; their  sum  will  be  the 
area  of  the  figure. 


EXAMPLE.— It  is  required  to  find  the  area  of  the  polygon  A  B  C  D  E  F, 
Fig.  37. 

SOLUTION. — Draw  the  diagonals  B  F  and  CF  and  the  line  FG 
perpendicular  to  D  £,  dividing  the  figure  into  the  triangles  A  B  F, 
B  C  F,  and  FG  E,  and  the  rectangle  FC  D  G.  Let  it  be  supposed  that 
the  altitudes  of  the  figures  and  the  lengths  of  the  sides  A  B,  D  G,  and 
G  E  are  as  indicated  in  the  polygon  above.  Then, 


Are&ABF    = 


16X7 


=  56  sq.  in. 


Area  FC  D  G  =  14  X  10  =  140  sq.  in. 
AreaFGE     =  9  ^   =45   sq.  in. 
Total  area  =  56  +  35  -f- 140  +  45  =  276  sq.  in.     Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  How  many  degrees  are  there  in  one  of  the  angles  of  a  regular 
octagon  ?  Ans.  135°. 

2.  Find  the  area  of  the  polygon  A  BCDEF  (see  Fig.  37),  sup- 
posing each  of  the  given  dimensions  to  be  increased  to  \\  times  the 
length  given  in  the  figure.  Ans.  621  sq.  in. 

3.  What  is  the  area  of  a  regular  heptagon,  whose  sides  are  4  inches 
long,  the  distance  from  one  side  to  the  center  being  4. 15  inches  ? 

Ans.  58.1  sq.  in. 

4.  At  what  angle  should  the  saw  run  in  a  miter-box  to  cut  strips  to 
fit  around  the  edge   of  a  table  top,  made  in  the  shape  of  a  regular 
pentagon?  Ans.  54°. 


MENSURATION. 


133 


THE    CIRCLE. 

392.     A     circle    (Fig.    38)   is   a    figure   / 
bounded  by  a  curved  line,  called  the  circum- 
ference, every  point  of  which  is  equally  dis- 
tant from  a  point  within,  called  the  center. 

s* \  FIG.  38. 

\  393.     The  diameter  of  a  circle  is  a 

J jg  straight   line  passing  through  the   center 

\  I     and   terminated  at  both  ends  by  the  cir- 

\  /       cumference;  thus,  A  B  (Fig.  39)  is  a  diam- 

eter of  the  circle. 


394.     The  radius  of  a  circle,  A  O  (Fig. 
40),  is  a  straight  line  drawn  from  the  center 
O    to   the   circumference.      It    is    equal    in 
length  to  one-half  the  diame- 
ter.     The  plural  of  radius  is 
radii,  and  all  radii  of  a  circle 
are  equal. 


FIG.  40. 


395.     An  arc  of  a  circle  (see  a  e  b,  Fig. 
41)  is  any  part  of  its  circumference. 


396.  A  chord  is  a  straight  line  joining 
any  two  points  in  a  circumference;  or  it  is  a 
straight  line  joining  the  extremities  of  an  arc ; 
thus,  the  straight  line  a  b,  Fig.  42,  is  a  chord 
of  the  circle  whose  corresponding  arc  is 
a  e  b. 


FIG.  42. 


397.     An    inscribed   angle   is   one 

whose  vertex  lies  on  the  circumference  of 
a  circle,  and  whose  sides  are  chords.  It 
is  measured  by  one-half  the  intercepted 
arc.  Thus,  in  Fig.  43,  A  B  C  is  an  in- 
c  scribed  angle,  and  it  is  measured  by  one- 
half  the  arc  ADC. 


134  MENSURATION. 

EXAMPLE. — If  in  Fig.  43,  the  arc  A  D  C  =  f  of  the  circumference, 
what  is  the  measurement  of  the  inscribed  angle  A  B  C? 

SOLUTION.— Since  the  angle  is  an  inscribed  angle,  it  is  measured  by 
one-half  the  intercepted  arc,  or  f  X  1  =  i  of  the  circumference.  The 
whole  circumference  =  360° ;  hence,  360°Xi  =  72°;  therefore,  angle 
A  B  C  is  an  angle  of  72°. 

398.  If  a  circle  is  divided  into  halves,  each  half  is  called 
a   semicircle,    and   each  half    circumference    is   called    a 

semi-circumference. 

Any  angle  inscribed  in  a  semicircle  is 
a  right  angle,  since  it  is  measured  by 
one-half  a  semi-circumference,  or  180° 
-=-2  =  90°.  Thus,  the  angles  A  D  C 
and  ABC,  Fig.  44,  are  right  angles, 
since  they  are  inscribed  in  a  semicircle. 

399.  An  inscribed  polygon  is  one  whose  vertexes  lie  on 
the  circumference  of  a  circle,  and  whose 

sides  are  chords,  as  A  B  C  D  E,  Fig.  45. 

The  sides  of  an  inscribed  regular  hex- 
agon have  the  same  length  as  the  radius 
of  the  circle. 

If,  in  any  circle,  a  radius  be  drawn  per- 
pendicular to  any  chord,  it  bisects  (cuts 
in  halves)  the  chord.  Thus,  if  the  radius  FlG  45- 

O  C,  Fig.  46,  is  perpendicular  to  the  chord  A  B,  A  D  =  D  B. 

A  EXAMPLE. — If  a  regular  pentagon  be  inscribed 

in  a  circle,  and  a  radius  is  drawn  perpendicular 
to  one  of  the  sides,  what  are  the  lengths  of  the 
two  parts  of  the  side,  the  perimeter  of  the  penta- 
gon being  27  inches  ? 

SOLUTION. — A  pentagon  has  five  sides,  and 
since  it  is  a  regular  pentagon,  all  the  sides  are  of 
equal  lengths;  the  perimeter  of  a  pentagon, 
which  equals  the  distance  around  it,  or  equals 
the  sum  of  all  the  sides,  is  27  inches.  There- 
fore, the  length  of  one  side  =  27  -=-  5  =  5f  inches.  Since  the  penta- 
gon is  an  inscribed  pentagon,  its  sides  are  chords,  and  as  a  radius 
perpendicular  to  a  chord  bisects  it,  we  have  51  -f-  2  =  2T7ff  inches,  which 
equals  the  length  of  each  of  the  parts  of  the  side,  cut  by  a  radius  per- 
pendicular to  it. 


MENSURATION.  135 

400.  If,  from  any  point  on  the  circumference  of  a  circle, 
a  perpendicular  be  let  fall  upon  a  given  diameter,  it  will  di« 
vide  the  diameter  into  two  parts,  one  of 

which  will  be  in  the  same  ratio  to  the  per- 
pendicular as  the  perpendicular  is  to  the 
other.    That  is,  the  perpendicular  will  be  a-4' 
mean  proportional  between  the  two  parts. 
If  A  B,  Fig.  47,  is  the  given  diameter, 
and  C  any  point  on  the  circumference, 
then  AD:  C  D:\  C  D:  D  B,  CD  being  a  mean  proportional 
between  A  D  and  D  B. 

EXAMPLE.— If  H K '=  30  feet,  and  I  fi  =  &  feet,  what  is  the  diameter 
of  the  circle,  H  K  being  perpendicular  to  A  B  ? 

SOLUTION.— 30  feet  -f-  2  feet  =  15  feet  =  I  H.  And  B I :  IH  ::  IH : 
I  A,  or  8  :  15  ::  15  :  I  A. 

Therefore,    I A  =  -^  =  ?P  =  28£  feet,  and  I A  +  IB  =  28i  -t-  8  = 

o  O 

36|  feet  =  A  B  the  diameter  of  the  circle.     Ans. 

401.  When  the  diameter  of  a  circle  and  the  lengths  of 
the  two  parts  into  which  it  is  divided  are  given,  the  length 
of  the  perpendicular  may  be  found  by  multiplying  the  lengths 
of  the  two  parts  together  and  extracting  the  square  root  of 
the  product. 

EXAMPLE.— In  Fig.  47,  the  diameter  of  the  circle  A  B  is  36£  feet, 
and  the  distance  B I  is  8  feet;  what  is  the  length  of  the  line  H  K  ? 

SOLUTION.— As  the  diameter  of  the  circle  is  36£  feet,  and  as  B I  is  8 
feet,  1 A  is  equal  to  36 \  —  8  =  28£  feet.  The  two  parts,  therefore,  are 

09^ 

8  and  28£  feet,  and  their  product  =  8  X  28£  =  8  X  -5-  =  225 ;  the  square 

O 

root  of  their  product  =  ^225  =  15  feet,  and  as  H  K—  IH+  IK,  or  2 
IH,  HK=  15  X  2  =  30  feet.     Ans. 

402.  To  find  the  circumference  of  a  circle,  the  diameter 
being  given: 

Rule  55. — Multiply  the  diameter  by  8.1416. 
EXAMPLE.— What  is  the  circumference  of  a  circle  whose  diameter  is 
15  inches  ? 

SOLUTION.— 15  X  3. 1416  =  47. 124  inches.     Ans. 

403.  To  find  the  diameter  of  a  circle,  the  circumference 
being  given : 


136  MENSURATION. 

Rule  56. — Divide  the  circumference  by  S.  1416. 

EXAMPLE.— What  is  ths  diameter  of  a  circle  whose  circumference  is 
65.973  inches? 

SOLUTION. — 65.973  -*-  3.1416  =  21  inches.     Ans. 

404.  To  find  the  length  of  an  arc  of  a  circle: 

Rule  57. — Multiply  the  length  of  the  circumference  of  the 
circle  of  which  the  arc  is  a  part  by  the  number  of  degrees  in 
the  arc,  and  divide  by  860. 

EXAMPLE.— What  is  the  length  of  an  arc  of  24°,  the  radius  of  the  arc 
being  18  in.  ? 

SOLUTION. — 18  X  2  =  36  in.  =  the  diameter  of  the  circle.  36  X 
3.1416  =  113.1  in.,  the  circumference  of  the  circle  of  which  the  arc  is  a 
part. 

24 

113.1  X  57^  =  7.54  in.,  or  the  length  of  the  arc.     Ans. 

ouu 

405.  To  find  the  area  of  a  circle : 

Rule  58. — Square  the  diameter,  and  multiply  by  .785 '4- 
EXAMPLE. — What  is  the  area  of  a  circle  whose  diameter  is  15  inches  ? 
SOLUTION.— 152  =  225;  and  225  X  .7854  =  176.72  sq.  in.     Ans. 

406.  Given  the  area  of  a  circle,  to  find  its  diameter: 
Rule  59. — Divide  the  area  by  .  7854,  and  extract  the  square 

root  of  the  quotient. 

EXAMPLE.— The  area  of  a  circle  =  17,671.5  square  inches.  What  is 
its  diameter  in  feet  ? 


SOLUTION.— |/17' 671'^  =  150  inches. 
r       . 7854 


.7854 

eet,  or  the  diameter.     Ans. 


EXAMPLE.—  What  is  the  area  of  a  flat  circular 
ring.  Fig.  48,  v/hose  outside  diameter  is  10  inches. 
and  whose  inside  diameter  is  4  inches  ? 

SOLUTION.—  The  area  of  the  large  circle  =  102  X 
.7854  =  78.54  square  inches;  the  area  of  the  small 
circle  =  42  X  -7854  =  12.57  square  inches.  The 
area  of  the  ring  is  the  difference  between  these 
areas,  or  78.54—  12.57  =  65.97  square  inches.  Ans. 

4O7.  To  find  the  area  of  a  sector  (a  sector  of  a  cir- 
cle is  the  area  included  between  two  radii  and  the  circum- 
ference, as,  for  example,  the  area  C  O  £,  Fig.  13): 


MENSURATION.  137 

Rule  6O.  —  Divide  the  number  of  degrees  in  the  arc  of  the 

sector  by  360.     Multiply  the  result  by  the  area  of  the  circle  of 
which  the  sector  is  a  part. 

EXAMPLE.  —  The  number  of  degrees  in  the  angle  formed  by  drawing 
radii  from  the  center  of  a  circle  to  the  extremities  of  the  arc  of  the 
circle  is  75°.  The  diameter  of  the  circle  is  12  inches;  what  is  the  area 
of  the  sector  ? 

SOLUTION.  —  ?-==  =  ^-j-;  and  12'2  X  .7854  =  113.1  square  inches. 

ODU          j&4 

113.1  X  HJ  =  23.56  square  inches,  the  area.     Ans. 

4O8.  To  find  the  area  of  a  segment  of  a  circle  (a 
segment  of  a  circle  is  the  area  included  between  a  chord 
and  its  arc;  for  example,  the  area  ABC,  Fig".  49): 

Rule  6  1  .  —  Divide  the  diameter  by  the  height  of  the  segment; 
subtract  .608  from  the  quotient,  and  extract  the  square  root  of 
the  remainder.  This  result  multiplied  by  4  times  the  square 
of  the  height  of  the  segment,  and  then  divided  by  3,  will  give 
the  area,  very  nearly. 

The  rule,  expressed  as  a  formula,  is  as  follows,  where 
D  =  the  diameter  of  the  circle,  and  h  =  the  height  of  the 
segment,  see  Fig.  49: 

Area  of  A  B  CA  =  A?  V=r  ~  •  608- 

O  II 

EXAMPLE.  —  What  is  the  area  of  the  segment 
of  a  circle  whose  diameter  is  54  inches,  the 
height  of  the  segment  being  20  inches  ? 

SOLUTION.  —  Substituting  in  the  formula, 

4X20*      54~~  4X20*      4X400 


=  4/2.092  =  1.447;   ^~-  X  1.447  =  771.7  sq.  in. 

Ans. 

EXAMPLES  FOR  PRACTICE. 

1.  An  angle  inscribed  in  a  circle  intercepts  one-third  of  the  circum- 
ference.    How  many  degrees  are  there  in  the  angle  ?  Ans.  60°. 

2.  Suppose   that   in  Fig.  47,  the  diameter  A  B  =  15  feet,  and  the 
distance  B  /=  3  feet.     What  is  the  length  of  the  line  HK  ? 

Ans.  12ft 


138 


MENSURATION. 


3.  The  diameter  of  a  fly-wheel  is  18  feet.     What  is  the  distance 
around  it  to  the  nearest  16th  of  an  inch  ?  Ans.  56  ft,  GT95  in. 

4.  A  carriage  wheel  was  observed  to  make  71|-  turns  while  going 
300  yards.     What  was  its  diameter  ?  Ans.  4  ft. ,  nearly. 

5.  What  is  the  length  of  an  arc  of  64°,  the  radius  of  the  arc  being 
30  inches?  Ans.  33.51  in. 

6.  Find  the  area  of  a  circle  2  feet  3  inches  in  diameter. 

Ans.  3.976  sq.  ft. 

7.  What  must  be  the  diameter  of  a  circle  to  contain  100  square 
inches?  Ans.  11.28  in. 

8.  Compute  the  area  of  a  segment,  whose  height  is  11  inches,  and 
the  radius  of  whose  arc  is  21  inches.  Ans.  289.04  sq.  in. 

9.  Find  the  area  of  a  flat  circular  ring  whose  outside  diameter  is 
12  inches  and  whose  inside  diameter  is  6  inches.  A.ns.  84.82  sq.  in. 


THE    PRISM   AND    CYLINDER. 

409.  A  solid,  or  body,  has  three  dimensions:  length, 
breadth,  and  thickness.     The  sides  which  enclose  it  are  called 
the  faces,  and  their  intersections  are  called  edges. 

410.  A  prism  is  a  solid  whose  ends  are  equal  and  par- 
allel polygons,  and  whose  sides  are  parallelograms,      Prisms 
take  their  names  from  the  form  of  their  bases.     Thus,  a  tri- 
angular prism  is  one  having  a  triangle  for  its  base;  a  hexag- 
onal prism  is  one  having  a  hexagon  for  its  base,  etc. 

41 1.  A  cylinder  is  a  body  of  uniform  diameter  whose 
ends  are  equal  parallel  circles. 

41 2.  A  parallelopipedon  (Fig. 
50)  is  a  prism  whose  bases  (ends)  are 
parallelograms. 

413.  A  cube  (Fig.  51)  is  a  prism 

.-,  whose  faces  and  ends   are  squares.          FIG  51 
All  the  faces  of  a  cube  are  equal. 

In  the  case  of  plane  figures,  we  have  had  to  do 
with  perimeters  and  areas.  In  the  case  of  solids,  we  have 
to  do  with  the  areas  of  their  outside  surfaces,  and  with 
their  contents  or  volumes. 


MENSURATION.  139 

414.  The  entire  surface  of  any  solid  is  the  area  of 
the  whole  outside  of  the  solid,  including  the  ends. 

The  convex  surface  of  a  solid  is  the  same  as  the  entire 
surface,  except  that  the  areas  of  the  ends  are  not  included. 

415.  A  unit  of  volume  is  a  cube  each  of  whose  edges 
is  equal  in  length  to  the  unit.     The  volume  is  expressed  by 
the  number  of  times  it  will  contain  a  unit  of  volume. 

Thus,  if  the  unit  of  length  is  1  inch,  the  unit  of  volume 
will  be  the  cube  whose  edges  each  measure  1  inch,  this  cube 
being  1  cubic  inch ;  and  the  number  of  cubic  inches  the  solid 
contains  will  be  its  volume.  If  the  unit  of  length  is  1  foot, 
the  unit  of  volume  will  be  1  cubic  foot,  etc.  Cubic  inch,  cubic 
foot,  and  cubic  yard  are  abbreviated  to  cu.  in.,  cu.  ft.,  and 
cu.  yd.,  respectively. 

Instead  of  the  word  volume,  the  expression  cubical 
contents  is  sometimes  used. 

416.  To  find  the  area  of   the  convex  surface  of  a  prism 
or  cylinder: 

Rule  62. — Multiply  the  perimeter  of  the  base  by  the  altitude. 

EXAMPLE.— A  block  of  marble  is  24  inches  long,  and  its  ends  are 
9  inches  square.  What  is  the  area  of  its  convex  surface  ? 

SOLUTION.— 9  X  4  =  36  =  the  perimeter  of  the  base;  36x24  =  864 
sq.  in.,  the  convex  area.  Ans. 

To  find  the  entire  area  of  the  outside  surface,  add  the  areas  of  the 
two  ends  to  the  convex  area.  Thus,  the  area  of  the  two  ends 
=  9  X  9  X  2  =  162  sq.  in. ;  864  -+-  162  =  1,026  sq.  in.  Ans. 

EXAMPLE.— How  many  square  feet  of  sheet  iron  will  be  required 
for  a  pipe  1±  feet  in  diameter  and  10  feet  long,  neglecting  the  amount 
necessary  for  lapping  ? 

SOLUTION.— The  problem  is  to  find  the  convex  surface  of  a  cylinder 
U  feet  in  diameter  and  10  feet  long.  The  perimeter,  or  circumference, 
of  the  base  =  1£  X  3.1416  =  1.5  X  3.1416  =  4.712  feet.  The  convex  sur- 
face =  4.712  X  10  =  47.12  sq.  ft.  of  metal.  Ans. 

417.  To  find  the  volume  of  a  prism  or  a  cylinder: 
Rule  63.— Multiply  the  area  of  the  base  by  the  altitude. 

EXAMPLE.— What  is  the  weight  of  a  length  of  wrought-iron  shaft- 
ing 16  feet  long  and  2  inches  in  diameter  ?  Wrought  iron  weighs  .28 
pound  per  cubic  Inch. 


140  MENSURATION. 

SOLUTION. — The  shaft  is  a  cylinder  16  feet  long.  The  area  of  one 
end,  or  the  base,  =  2*  X  .7854  =  3.1416  sq.  in.  Since  the  weight  of  the 
iron  is  given  per  cubic  inch,  the  contents  of  the  shaft  must  be  found 
in  cubic  inches.  The  length,  16  feet,  reduced  to  inches  =  16x12  = 
192  in.;  3.1416x192  =  603.19  cu.  in.  =  the  volume.  The  weight  = 
603.19  X- 28  =  168.89  Ib.  Ans. 

EXAMPLE.— Find  the  cubical  contents  of  a  hexagonal  prism,  Fig.  52, 
12  inches  long,  each  edge  of  the  base  being  one  inch  long. 

SOLUTION. — In  order  to  obtain  the  area  of  one 
end,  the  distance  CD  from  the  center  Cto  one  side 
must  be  found. 

In  the  right-angled  triangle  C D  A,  side  A  D  = 
i  A  B,  or  one-half  inch,  and  since  the  polygon  is  a 
hexagon,  side  C  A  =  distance  A  B,  or  one  inch. 
Hence,  C  A  being  the  hypotenuse,  the   length  of 
FIG.  52.  side   C  D  =  /I*  -  (i)*  =  4/1*  -  .5*  =  1/7*75.  or 

.866  inch.  Area  of  triangle  A  CS  =  *  x^866  =  .433  sq.  in. .  area  of  the 
whole  polygon  =  .433  X  6  =  2.598  sq.  in.  Hence,  the  contents  of  the 
prism  =  2.598  X  12  =  31.176  cu.  in.  Ans. 

EXAMPLE. — It  is  required  to  find  the  number  of  cubic  feet  of  steam 
space  in  the  boiler  shown  in  Fig.  53,  The  boiler  is  16  feet  long  between 


FIG.  53. 

heads,  54  inches  in  diameter,  and  the  mean  water  line  M  N'\s  at  a  dis- 
tance of  16  inches  from  the  top  of  the  boiler.  The  volume  of  the  steam 
outlet  casting  may  be  neglected. 

SOLUTION. — The  volume  of  the  steam  space,  which  is  that  space 
within  the  boiler  above  the  surface  M NO  P  of  the  water,  is  found  by 
the  rule  for  finding  the  volume  of  a  prism  or  cylinder,  the  area  M ' N  S 
being  the  base,  and  the  length  'N  O  the  altitude.  First  obtain  the  area 
of  the  segment  M  N  S,  whose  height  h  is  16  inches,  in  square  feet;  then 
multiply  the  result  by  16,  the  length  of  the  boiler. 

By  the  formula  previously  given,  the  area  of  the  segment  = 


MENSURATION.  141 


-  .608  =  4/27767  =  1.663. 

Hence,  the  area  =  341.33  X  1.663  =  567.63  sq.  in.  This,  reduced  to 
square  feet,  =  567.63  -H  144  =  3.942sq.  ft.,  and  the  volume,  therefore,  = 
3.942  x  16  =  63.07  cu.  ft.  Ans. 

In  the  above  solution,  the  space  occupied  by  the  stays  is 
not  considered,  for  sake  of  simplicity.  They  are  not  shown 
in  the  figure. 

EXAMPLE.—  In  the  above  boiler  there  are  60  tubes,  3J  inches  outside 
diameter.  How  many  gallons  of  water  will  it  take  to  fill  the  boiler  up 
to  the  mean  water  level,  there  being  231  cubic  inches  in  a  gallon  ? 

SOLUTION.  —  Find  the  volume  in  cubic  inches  of  that  part  of  the  boiler 
below  the  surface  of  the  water  M  NO  P,  since  the  contents  of  a  gallon  is 
given  in  cubic  inches,  and  from  it  subtract  the  volume  of  the  tubes  in 
cubic  inches. 

This  may  be  done  by  first  finding  the  to/at  area.  of  one  end  of  the  boiler 
in  square  inches,  from  it  subtracting  the  area  of  the  segment  M  N  S,  and 
the  areas  of  the  ends  of  the  tubes  in  square  inches,  and  then  by  multi- 
plying the  result  by  the  length  of  the  boiler  in  inches. 

Total  area  of  one  end  =  542  X  .7854  =  2,290.23  sq.  in. 

Area  of  segment  M  NS,  as  found  in  last  example,  =  567.63  sq.  in. 

Area  of  the  end  of  one  tube  =  3.252  X  .7854  —  8.2958  sq.  in. 

Area  of  the  ends  of  the  60  tubes  =  8.2958  X  60  =  497.75  sq.  in. 

Hence,  the  area  to  be  subtracted  =  567.63  +  497.75  =  1,065.38  sq.  in. 
Subtracting,  2,290.23  -  1,065.38  =  1,224.85  sq.  in.  =  net  area. 

The  cubical  contents  =  1,224.85  X  16  X  12  =  235,171.2  cu.  in.  This, 
divided  by  231,  will  give  the  number  of  gallons;  whence,  235,171.2-*- 
231  =  1,018.06  gallons  of  water.  Ans. 


EXAMPLES   FOR   PRACTICE. 

1.  Find  the  area. in  square  inches  of  the  convex  surface  of  a  bar  of 
iron  4£  inches  in  diameter,  and  8  feet  5  inches  long.    Ans.  1,348. 53  sq.  in. 

2.  Find  the  area  of  the  entire  surface  of  the  above  bar. 

Ans.  1,376.9  sq.  in. 

3.  What  is  the  area  of  the  entire  surface  of  the  hexagonal  prism 
whose  base  is  shown  in  Fig.  52  ?  Ans.  77.196  sq.  in. 

4.  A  multitubular  boiler  has  the  following  dimensions:  diameter,  50 
inches;  length  between  heads,  15  feet;  number  of  tubes,  56;  outside 
diameter  of  tubes,  3  inches;  distance  of  mean  water  line  from  top  of 
boiler,  16  inches,     (a)  Compute  the  steam  space  in  cubic  feet.     (A)  Find 
the  number  of  gallons  of  water  required  to  fill  the  boiler  up  to  the  mean 
water  line.  .        j  (a)  56.4  cu.  ft. 

ADS'  (  (6)  800  gallons. 


143 


MENSURATION. 


THE    PYRAMID    AND    CONE. 

418.  A    pyramid  (Fig.  54)  is   a  solid 
whose  base  is  a  polygon,  and  whose  sides  are 
triangles  uniting  at  a  common 

point,  called  the  vertex. 

419.  A  cone  (Fig.  55)  is 
a  solid  whose  base  is  a  circle 
and     whose    convex    surface 

tapers  uniformly  to  a  point  called  the  vertex. 

420.  The  altitude  of  a  pyramid  or  cone 

is  the  perpendicular  distance  from  the  vertex  to  the  base. 

421.  The  slant  height  of  a  pyramid  is  a  line  drawn 
from   the  vertex  perpendicular  to  one  of  the  sides  of  the 
base.     The  slant  height  of  a  cone  is  any  straight  line  drawn 
from  the  vertex  to  the  circumference  of  the  base. 

422.  To  find  the  convex  area  of  a  pyramid  or  cone: 
Rule  64. — Multiply  the  perimeter  of  the  base  by  one-half 

the  slant  height. 

EXAMPLE. —What  is  the  convex  area  of  a  pentagonal  pyramid,  if  one 
side  of  the  base  measures  6  inches,  and  the  slant  height  =  14  inches  ? 

SOLUTION.— The  base  of  a  pentagonal  pyramid  is  a  pentagon,  and, 
consequently,  has  five  sides. 

6  x  5  =  30  inches,  or  the  perimeter  of  the  base.  30  X  -g-  =  21°  SQ-  in-> 
or  the  convex  area.  Ans. 

EXAMPLE. — What  is  the  entire  area  of  a  right  cone  whose  slant 
height  is  1?  inches,  and  whose  base  is  8  inches  in  diameter  ? 

SOLUTION.— The  perimeter  of  the  base  =  8x3. 1416  =  25.1328  in. 

Convex  area  =  25.1328  X  ^  =  213.63  sq.  in. 
Area  of  base  =       82  X  .7854  =    50.27  sq.  in. 


Entire  area  =  263.90  sq.  in.     Ans. 

423.     To  find  the  volume  of  a  pyramid  or  cone: 
Rule  65. — Multiply  the  area  of  the  base  by  one-third  of 
the  altitude. 

EXAMPLE. — What  is  the  volume  of  a  triangular  pyramid,  each  edge 
of  whose  base  measures  6  inches,  and  whose  altitude  is  8  inches? 


MENSURATION. 


143 


SOLUTION.— Draw  the  base  as  shown  in  Fig.  56; 
it  will  be  an  equilateral  triangle,  all  of  whose 
sides  are  6  inches  long. 

Draw  a  perpendicular  B  D  from  the  vertex 
to  the  base;  it  will  divide  the  base  into  two 
equal  parts,  since  an  equilateral  triangle  is  also 
isosceles,  and  will  be  the  altitude  of  the  triangle. 
In  order  to  obtain  the  area  of  the  base,  this 
altitude  must  be  determined. 

In  the  right-angled  triangle  B  D  A,  the  hypotenuse  B  A  =  6  inches, 
and  side  A  D  =  3  inches,  to  find  the  other  side, 

B  D—  4/6*  —  3*  —  5.2  inches,  nearly. 

Area  of  the  base,  or  B  A  C,  =  —     —  =  15.6  sq.  in.    Hence,  the  volume 


D 

FIG.  50. 


=  15.6  x|-  =  41.6  cu.  in. 
o 


Ans. 


EXAMPLE. — What  is  the  volume  of  a  cone  whose  altitude  is  18 
inches,  and  whose  base  is  14  inches  in  diameter  ? 

SOLUTION.— Area  of  the  base  =  14*  X  .7854  =  153.94  sq.  in.  Hence, 
the  volume  =  153.94  X -5- =  923.64  cu.  in.  Ans. 

O 

EXAMPLES   FOR   PRACTICE. 

1.  Find  the  convex  surface  of  a  square  pyramid  whose  slant  height 
is  28  inches,  and  one  edge  of  whose  base  is  7£  inches  long. 

Ans.  420  sq.  in. 

2.  What  is  the  volume  of  a  triangular  pyramid,  one  edge  of  whose 
base  measures  3  inches,  and  whose  altitude  is  4  inches  ? 

Ans.  5.2  cu.  in. 

3.  Find  the  volume  of  a  cone  whose  altitude  is  12  inches,  and  the 
circumference  of  whose  base  is  31.416  inches.  Ans.  314.16  cu.  in. 

NOTE. — Find  the  diameter  of  the  base  and  then  its  area. 


THE    FRUSTUM    OF    A    PYRAMID    OR    COXE. 


424.  If  a  pyramid  be  cut  by  a 
plane,  parallel  to  the  base,  so  as 
to  form  two  parts,  as  in  Fig.  57, 
the  lower  part  is  called  the  frustum 
of  the  pyramid. 

If  a  cone  be  cut  in  a  similar  man- 
ner, as  in  Fig.  58,  the  lower  part  is 
called  the  frustum  of  the  cone. 


Flo.  58. 


144  MENSURATION. 

425.  The  upper  end  of  the  frustum  of  a  pyramid  or 
cone    is    called    the    upper     base,  and  the  lower  end  the 
lower  base.     The  altitude  of  a  frustum  is  the  perpendicu- 
lar distance  between  the  bases. 

426.  To  find  the  convex  surface  of  a   frustum   of   a 
pyramid  or  cone: 

Rule  66.  —  Multiply  onc-Jialf  the  sum  of  tlie  perimeters  of 
tJie  two  bases  by  the  slant  heigJit  of  tJie  frustum. 

EXAMPLE.  —  Given,  the  frustum  of  a  triangular  pyramid,  in  which 
one  side  of  the  lower  base  measures  10  inches,  one  side  of  the  upper 
base  measures  6  inches,  and  whose  slant  height  is  9  inches  ;  find  the 
area  of  the  convex  surface. 

SOLUTION.  —  10  in.  X  3  =  30  in.,  the  perimeter  of  the  lower  base. 

6  in.  X  3  =  18  in.,  the  perimeter  of  the  upper  base. 

-  2  -  =  24  in.,  or  one-half  the  sum  of  the  perimeters  of  the  two 
bases.  24  X  9  =  216  sq.  in.,  the  convex  area.  Ans. 

EXAMPLE.  —  If  the  diameters  of  the  two  bases  of  a  frustum  of  a  cone 
are  12  inches  and  8  inches,  respectively,  and  the  slant  height  is  12 
inches,  what  is  the  entire  area  of  the  frustum  ? 


SOLUTION.-  -  ><  -  X  18  =  376.99  sq.  in.,  the 

area  of  the  convex  surface. 

Area  of  the  upper  base  =    8'2  X  .7854  =  50.27  sq.  in. 

Area  of  the  lower  base  =  12*  X  .7854  =  113.1  sq.  in. 

The  entire  area  of  the  frustum  =  376.99  +  50.27  +  113.1  =  540.30 
1  -•[.  in.  Ans. 

427.  To  find  the  volume  of  the  frustum  of  a  pyramid 
or  cone: 

Rule  67.  —  Add  together  the  att>as  of  the  upper  and  loivcr 
bases,  and  the  square  root  of  tJie  product  of  the  two  areas  ; 
multiply  the  sum  by  one-third  of  the  altitude. 

EXAMPLE.  —  Given,  a  frustum  of  a  square  pyramid  (one  whose  base 
is  a  square)  ;  each  edge  of  the  lower  base  is  12  inches,  each  edge  of  the 
upper  base  is  5  inches,  and  its  altitude  is  16  inches;  what  is  its  volume  ? 

SOLUTION.—  Area  of  upper  base  =  5  X  5  =  25  sq.  in.  ;  area  of  lower 
base  =  12  X  12  =  144  sq.  in.  ;  the  square  root  of  the  product  of  the  two 
areas  =  4/25  X  144  =  60.  Adding  these  three  results,  and  multiplying 

by  one-third    the  altitude,    25  +  144  +  60  =  229;  229  X  4r  =  1.22H  cu. 

o 

in.  =  the  volume.     Ans. 


MENSURATION.  145 

EXAMPLE.— How  many  gallons  of  water  will  a  round  tank  hold, 
which  is  4  feet  in  diameter  at  the  top,  5  feet  in  diameter  at  the  bottom, 
and  8  feet  deep  ? 

SOLUTION.— There  are  231  cubic  inches  in  a  gallon,  and  the  volume 
of  the  tank  should  be  found  in  cubic  inches.  The  tank  is  in  the  shape 
of  the  frustum  of  a  cone.  The  upper  diameter  =  4  X  12  =  48  inches; 
the  lower  diameter  =  5  X  12  =  60  inches,  and  the  depth  =  8  X  12  =  96 
inches.  Area  of  upper  base  =  48*  X  .7854  =  1,809.56  sq.  in. ;  area  of  lower 
base  =  60*  X  .7854  =  2,827. 44 sq.  in. ;  4/1,809. 56  X  2,827.44  =  2,261.95. 

O/» 

Whence,    1,809.56  +  2,827.44  +  2,261.95  =  6,898.95;    6,898.95  X  ^r  = 

o 

220,766.4  cu.  in.  =  contents.     Now,  since  there  are  231  cu.  in.  in  one 
gallon,  the  tank  will  hold  220,766.4  -*-  231  =  955.7  gallons,  nearly.   Ans. 


EXAMPLES   FOR  PRACTICE. 

1.  Find  the  convex  surface  of  the  frustum  of  a  square  pyramid,  one 
edge  of  whose  lower  base  is  15  inches  long,  one  edge  of  whose  upper 
base  is  14  inches  long,  and  whose  slant  height  is  one  inch. 

Ans.  58  sq.  in. 

2.  Find  the  volume  of  the  above  frustum,  supposing  its  altitude  to 
be  3  inches.  Ans.  631  cu.  in. 

3.  Find  the  volume  of  the  frustum  of  a  cone  whose  altitude  is  12 
feet  and  the  diameters  of  whose  upper  and  lower  bases  are  8  and  10  feet, 
respectively.  Ans.  766.55  cu.  ft. 

4.  If  a  tank  had  the  dimensions  of  example  3,  how  many  gallons 
would  it  hold  ?  Ans.  5,734.2  gallons,  nearly. 

THE    SPHERE   AND    CYLINDRICAL    RING. 

428.  A    sphere    (Fig.    59)    is    a   solid 
bounded    by   a   uniformly   curved   surface, 
every  point  of  which  is  equally  distant  from 
a  point  within,  called  the  center. 

The   word    ball,  or  globe,   is   generally 
used  instead  of  sphere. 

429.  To  find  the  area  of  the  surface  of  a 
sphere : 

Rule  68. — Square  the  diameter  and  multiply  the  result 
by  3. 1416. 

EXAMPLE.— What  is  the  area  of  the  surface  of  a  sphere  whose  diam- 
eter is  14  inches  ? 

SOLUTION.— Diameter  squared  X  3.1416  =  14s  X  3.1416  =  14  X  14  X 
3.1416  =  615.75  sq.  in.  Ans. 


146  MENSURATION. 

From  this  it  will  be  seen  that  the  surface  of  a  sphere 
equals  the  circumference  of  a  great  circle  multiplied  by  the 
diameter,  a  rule  often  used ;  a  great  circle  of  a  sphere  is  the 
intersection  of  its  surface  with  a  plane  passing  through  its 
center;  for  instance,  the  great  circle  of  a  sphere  6  in. 
diameter  is  a  circle  of  6  in.  diameter.  Any  number  of 
great  circles  could  be  described  on  a  given  sphere. 

43O.     To  find  the  volume  of  a  sphere : 

Rule  69.- — Cube  the  diameter  and  multiply  the  result  by 


EXAMPLE. — What  is  the  weight  of  a  lead  ball  12  inches  in  diameter, 
a  cubic  inch  of  lead  weighing  .41  pound  ? 

SOLUTION.  — Diameter  cubed  X  .5236  =  12  X  12  X  12  X  .5236  =  904.78 
cu.  in.,  or  the  volume  of  the  ball.  The  weight,  therefore,  =  904.78  X 
.41  =  370.96  pounds.  Ans. 

431.     To  find  the  convex  area  of  a  cylindrical  ring: 

A  cylindrical  ring  (Fig.  60)  is  a  cyl- 
inder bent  to  a  circle.  The  altitude  of  the 
cylinder  before  bending  is  the  same  as 
the  length  of  the  dotted  center  line  D. 
The  base  will  correspond  to  a  cross- 
section  on  the  line  A  B  drawn  from  the 
center  O.  Hence,  to  find  the  convex 
area: 

Rule  7O. — Multiply  the  circumference  of  an  imaginary 
cross-section  on  the  line  A  B,  by  the  length  of  the  center 
line  D. 

EXAMPLE. — If  the  outside  diameter  of  the  ring  is  12  inches,  and  the 
inside  diameter  is  8  inches,  what  is  its  convex  area  ? 

SOLUTION. — The  diameter  of  the  center  circle  equals  one-half  the 

-|  (>        -       Q 

sum   of    the  inside  and    outside    diameters  =  — 5—  =  10,   and   10  X 

3.1416  =  31.416  inches,  the  length  of  the  center  line. 

The  radius  of  the  inner  circle  is  4  inches;  of  the  outside  circle, 
6  inches ;  therefore,  the  diameter  of  the  cross-section  on  the  line  A  B 
is  2  inches.  Then,  2  x  3.1416  =  6.2832  inches,  and  6.2832x31.416- 
197.4  sq.  in.,  the  convex  area.  Ans. 


MENSURATION.  U\ 

432.     To  find  the  volume  of  a  cylindrical  ring: 
Rule  71. —  The  volume  will  be  the  same  as  that  of  a  cylin- 
der  whose  altitude  equals  the  length  of 
the  dotted  center  line  D,  and  whose  base 
is  the  same  as  a  cross-section  of  the  ring 
on  the  line  A  B,  drawn  from  the  center 
O.     Hence,  to  find  the  volume  of  a  cylin- 
drical   ring,    multiply   the    area   of   an 
imaginary  cross-section  on  the  line  A  B, 
by  the  length  of  the  center  line  D.  FIG.  «i. 

EXAMPLE. — What  is  the  volume  of  a  cylindrical  ring  whose  outside 
diameter  is  12  inches,  and  whose  inside  diameter  is  8  inches  ? 

SOLUTION.— The  diameter  of  the  center  circle  equals  one-half  the 

•I  Q       ,       Q 

sum  of  the  inside  and  outside  diameters  =  — ^ —  =  10. 

10  X  3.1416  =  31.416  inches,  the  length  of  the  center  line. 

The  radius  of  the  outside  circle  =  6  inches;  of  the  inside  circle  = 
4  inches;  therefore,  the  diameter  of  the  cross-section  on  the  line  A  B  = 
2  inches. 

Then,  2*  x  .7854  =  3.1416  sq.  in.,  the  area  of  the  imaginary  cross- 
section. 

And  3.1416  X  31.416  =  98.7  cubic  inches,  the  volume.     Ans. 


EXAMPLES  FOR   PRACTICE. 

1.     What  is  the  volume  of  a  sphere  30  inches  in  diameter  ? 

Ans.   14, 137. 2  cu.  in. 
2.     How  many  square  inches  in  the  surface  of  the  above  sphere  ? 

Ans.  2,827.44  sq.  in. 

3.  Required  the  area  of  the  convex  surface  of  a  circular  ring,  the 
outside  diameter  of  the  ring  being  10  inches,  and  the  inside  diameter 
74  Inches.  Ans.  107.95  sq.  in. 

4.  Find  the  cubical  contents  of  the  ring  in  the  last  example. 

Ans.  33.73cu.  in. 

5.  The  surface  of  a  sphere  contains  314.16  square  inches.     What  is 
the  volume  of  the  sphere  ?  Ans.  523.6  cu.  in. 


MECHANICS. 


433.  Mechanics  is  that  science  which  treats  ot  the 
action  of  forces  upon  bodies,  and  the  effects  which   they 
produce ;  it  treats  of  the  laws  which  govern  the  movement 
and   equilibrium   of   bodies,  and   shows  how   they  may  be 
utilized. 

MATTER  AND  ITS   PROPERTIES. 

434.  Matter  is  anything  that   occupies  space.     It   is 
the  substance  of  which  all  bodies  consist.     Matter  is  com- 
posed of  molecules  and  atoms. 

435.  A   molecule  is  the  smallest   portion   of   matter 
that  can  exist  without  changing  its  nature. 

436.  An  atom  is  an  indivisible  portion  of  matter. 
Atoms  unite  to  form  molecules,  and  a  collection  of  mole- 
cules forms  a  mass  or  body. 

A  drop  of  water  may  be  divided  and  subdivided,  until 
each  particle  is  so  small  that  it  can  only  be  seen  by  the 
most  powerful  microscope,  but  each  particle  will  still  be 
water. 

Now,  imagine  the  division  to  be  carried  on  still  further, 
until  a  limit  is  reached  beyond  which  it  is  impossible  to  go 
without  changing  the  nature  of  the  particle.  The  particle 
of  water  is  now  so  small  that,  if  it  be  divided  again,  it  will 
cease  to  be  water,  and  will  be  something  else;  we  will  call 
this  particle  a  molecule. 

If  a  molecule  of  water  be  divided,  it  will  yield  two  atoms 
of  hydrogen  gas,  and  one  of  oxygen  gas.  If  a  molecule 
of  sulphuric  acid  be  divided,  it  will  yield  two  atoms  of 
hydrogen,  one  of  sulphur,  and  four  of  oxygen. 


150  MECHANICS. 

It  has  been  calculated  that  the  diameter  of  a  molecule  is 
larger    than    TSTrroVoooo    °f    an    inch,   and    smaller    than 
of  an  inch- 


437.  Bodies  are  composed  of  collections  of  molecules. 
Matter  exists  in  three  conditions  or  forms:  solid,  liqtiid,  and 
gaseous. 

438.  A   solid    body   is   one  whose    molecules  change 
their  relative  positions  with  great  difficulty;  as  iron,  wood, 
stone,  etc. 

439.  A  liquid  body  is  one  whose  molecules  tend  to 
change  their  relative  positions  easily.      Liquids  readily  adapt 
themselves  to  the  shape  of  vessels  which  contain  them,  and 
their  upper  surface  always  tends  to  become  perfectly  level. 
Water,  mercury,  molasses,  etc.,  are  liquids. 

440.  A  gaseous  body,  or  gas,  is  one  whose  molecules 
tend    to    separate    from    one    another;    as    air,     oxygen, 
hydrogen,  etc. 

Gaseous  bodies  are  sometimes  called  aeriform  (air-like) 
bodies.  They  are  divided  into  two  classes:  the  so-called 
"  permanent"  gases,  and  vapors. 

441.  A  permanent  gas  is  one  which  remains  a  gas 
at  ordinary  temperatures  and  pressures. 

442.  A  vapor  is  a  body  which  at  ordinary  tempera- 
tures is  a  liquid  or  solid,  but  when  heat  is  applied,  becomes 
a  gas,  as  steam. 

One  body  may  be  in  all  three  states;  as,  for  example, 
mercury,  which  at  ordinary  temperatures  is  a  liquid, 
becomes  a  solid  (freezes)  at  40°  below  zero,  and  a  vapor 
(gas)  at  600°  above  zero.  By  means  of  great  cold,  all  gases, 
even  hydrogen,  have  been  liquefied,  and  some  solidified. 

By  means  of  heat,  all  solids  have  been  liquefied,  and  a 
great  many  vaporized.  It  is  probable  that,  if  we  had  the 
means  of  producing  sufficiently  great  extremes  of  heat  and 
cold,  all  solids  might  be  converted  into  gases,  and  all  gases 
into  solids. 


MECHANICS.  151 

443.  Every  portion  of  matter  possesses  certain  qualities 
called  properties.     Properties   of    matter   are    divided    into 
two  classes :  general  and  special. 

444.  General  properties  of  matter  are  those  which 
are  common  to  all  bodies.      They  are  as  follows:     Extension, 
impenetrability,  weight,  indestructibility,   inertia,   mobility, 
divisibility,    porosity,     compressibility,     expansibility,     and 
elasticity. 

445.  Extension  is  the  property  of  occupying  space. 
Since  all  bodies  must  occupy  space,  it  follows  that  extension 
is  a  general  property. 

446.  By  impenetrability  we  mean  that  no  two  bodies 
can  occupy  exactly  the  same  space  at  the  same  time. 

447.  Weight  is  the  measure  of  the  earth's  attraction 
upon  a  body.     All  bodies  have  weight.      In  former  times  it 
was  supposed  that  gases  had  no  weight,  since,  if  unconfined, 
they  tend  to  move  away  from  the  earth,  but,  nevertheless, 
they  will  finally  reach  a  point  beyond  which  they  can  not  go, 
being  held  in  suspension  by  the  earth's  attraction.      Weight 
is  measured  by  comparison  with  a  standard.     The  standard 
is  a  bar  of  platinum  owned  and  kept  by  the  Government;  it 
weighs  one  pound. 

448.  Inertia  means  that  a  body  can  not  put  itself  in 
motion  nor  bring  itself  to  rest.     To  do  either  it  must  be 
acted  upon  by  some  force. 

449.  Mobility  means  that  a  body  can  be  changed  in 
position  by  some  force  acting  upon  it. 

450.  Divisibility    is    that    property    of    matter    on 
account  of  which  a  body  may  be  separated  into  parts. 

451.  Porosity  is  the  term  used  to  denote  the  fact  that 
there  is  space  between  the  molecules  of  a  body.     The  mole- 
cules of  a  body  are  supposed  to  be  spherical,  and,  hence, 
there  is  space  between  them,  as  there  would  be  between 
peaches  in  a  basket.     The   molecules  of  water  are    larger 
than  those  of  salt;  so  that  when  salt  is  dissolved  in  water, 


152  MECHANICS. 

its  molecules  wedge  themselves  between  the  molecules  of  the 
water,  and  unless  too  much  salt  is  added,  the  water  will 
occupy  no  more  space  than  it  did  before.  This  does  not 
prove  that  water  is  penetrable,  for  the  molecules  of  salt 
occupy  the  space  that  the  molecules  of  water  did  not. 

Water   has   been  forced  through  iron  by  pressure,  thus 
proving  that  iron  is  porous. 

452.  Compressibility. — This   property  is   a   natural 
consequence   of    the  preceding  one.      Since  there  is  space 
between  the  molecules,  it  is  evident  that  by  means  of  force 
(pressure)  they  can  be  brought  closer  together,  and  thus  the 
body  be  made  to  occupy  a  smaller  space. 

453.  Expansibility  is  the   term  used  to  denote  the 
fact  that  the  molecules  of  a  body  will,  under  certain  condi- 
tions (when  heated,  for  example),  move  farther  apart,  and  so 
cause  the  body  to  expand,  or  occupy  a  greater  space. 

454.  Elasticity  is  that  property  of  matter  which  en- 
ables a  body  when  distorted  within  certain  limits  to  resume 
its  original   form  when   the   distorting  force   is   removed. 
Glass,  ivory,  and  steel  are  very  elastic,  clay  and  putty  in 
their  natural  state  being  very  slightly  so. 

455.  Indestructibility  is  the  term  used  to  denote  the 
fact  that  we  can  not  destroy  matter.     A  body  may  undergo 
thousands  of  changes,   be  resolved  into  its  molecules,  and 
its  molecules  into  atoms,  which  may  unite  with  other  atoms 
to  form  other  molecules  and  bodies  entirely  different  in  ap- 
pearance and  properties  from  the  original  body,  but  the  same 
number  of  atoms  remain.    The  whole  number  of  atoms  in  the 
universe  is  exactly  the  same  now  as  it  was  millions  of  years 
ago,  and  will  always  be  the  same.     Matter  is  indestmctible. 

456.  Special  properties  are  those  which  are  not  pos- 
sessed by  all  bodies.     Some  of  the  most  important  are  as 
follows:    Jiardness,     tenacity,    brittleness,     malleability,  and 
ductility. 

457.  Hardness. — A  piece   of    copper   will  scratch   a 
piece  of  wood,  steel  will  scratch  copper,  and  tempered  steel 


MECHANICS.  153 

will  scratch  steel  in  its  ordinary  state.  We  express  all  this 
by  saying  that  steel  is  Iiardcr  than  copper,  and  so  on.  Emery 
and  corundum  are  extremely  hard,  and  the  diamond  is  the 
hardest  of  all  known  substances.  It  can  only  be  polished 
with  its  own  powder. 

458.  Tenacity  is  the  term  applied  to  the  power  with 
which  some  bodies  resist  a  force  tending  to  pull  them  apart. 
Steel  is  very  tenacious. 

459.  Brittleness. — Some  bodies  possess  considerable 
power  to  resist  either  a  pull  or  a  pressure,  but  they  are  easily 
broken  when  subjected  to  shocks  or  jars;  for  example,  good 
glass  will  bear  a  greater  compressive  force  than  most  woods, 
but  may  be  easily  broken  when  dropped  on  to  a  hard  floor; 
this  property  is  called  brittleness. 

460.  Malleability  is  that  property  which  permits  of 
some  bodies  being  hammered  or  rolled  into  sheets.     Gold  is 
the  most  malleable  of  all  substances. 

461.  Ductility  is  that  property  which  enables  some 
bodies  to  be  drawn  into  wire.      Platinum  is  the  most  ductile 
of  all  substances. 


MOTION    A1VD    VELOCITY. 

462.  Motion  is  the  opposite  of    rest,   and  indicates  a 
changing  of  position  in  relation  to  some  object  which  is  for 
that  purpose  regarded  as  being  fixed.     If  a  large  stone  is 
rolled  down  hill,  it  is  in  motion  in  relation  to  the  hill. 

If  a  person  is  on  a  railway  train,  and  walks  in  the  oppo- 
site direction  from  that  in  which  the  train  is  moving,  and 
with  the  same  speed,  he  will  be  in  motion  as  regards  the 
train,  but  at  rest  with  respect  to  the  earth,  since,  until  he 
gets  to  the  end  of  the  train,  he  will  be  directly  over  the 
spot  at  which  he  was  when  he  started  to  walk. 

463.  The  path  of  a  body  in  motion  is  the  line  described 
by  a  certain  point  in  the  body  called  its  center  of  gravity. 
No  matter  how  irregular  the  shape  of  the  body  may  be,  nor 
how  many  turns  and  twists  it  may  make,  the  line  which 


154  MECHANICS. 

indicates  the  direction  of  this  point  for  every  instant  that  it 
is  in  motion  is  the  path  of  the  body. 

464.  Velocity  is  rate  of  motion.     It  is  measured  by  a 
unit  of  space  passed  over  in  a  unit  of  time.     When  equal 
spaces  are  passed  over  in  equal  times,  the  velocity  is  said  to 
be  uniform.     In  all  other  cases  it  is  variable. 

If  the  fly-wheel  of  an  engine  keeps  up  a  constant  speed  of 
a  certain  number  of  revolutions  per  minute,  the  velocity  of 
any  point  is  uniform.  A  railway  train  having  a  constant 
speed  of  40  miles  per  hour,  moves  40  miles  every  hour,  or 

40      2     , 

—  =  —  of  a  mile  every  minute,  and  since  equal  spaces  are 

passed  over  in  equal  times,  the  velocity  is  uniform. 

465.  To  find  the  uniform  velocity  which  a  body  must 
have  to  pass  over  a  certain  distance  or  space  in  a  given  time : 

Rule    72. — Divide  the  distance  by  the  time. 

EXAMPLE.— The  piston  of  a  steam  engine  travels  3,000  feet  in  5 
minutes;  what  is  its  velocity  in  feet  per  minute ? 

SOLUTION. — Here  3,000  feet  is  the  distance,  and  5  minutes  is  the  time. 
Applying  the  rule,  3,000  -r-  5  =  600  feet  per  minute.  Ans. 

CAUTION. — Before  applying  the  above  or  any  of  the  suc- 
ceeding rules,  care  must  be  taken  to  reduce  the  values  given 
to  the  denominations  required  in  the  answer.  Thus,  in  the 
above  example,  had  the  velocity  been  required  in  feet  per 
second  instead  of  feet  per  minute,  the  5  minutes  would  have 
been  reduced  to  seconds  before  dividing.  The  operation 
would  then  have  been,  5  min.  =  5  X  60  =  300  sec.  Applying 
the  rule,  3,000  -=-  300  =  10  ft.  per  sec.  Ans. 

466.  Had  the  velocity  been  required  in  inches  per  sec- 
ond, it  would  have  been  necessary  to  reduce  the  3,000  feet 
to  inches  and  the  5   minutes  to  seconds,  before  dividing. 
Thus,   3,000  ft.  X  12  =  36,000  in.      5  min.  X  60  =  300  sec. 
Now,  applying  the  rule,  36,000  -4-  300  =  120  in.  per  sec.    Ans. 

EXAMPLE. — A  railroad  train  travels  50  miles  in  l£  hours;  what  is  its 
average  velocity  in  feet  per  second  ? 


MECHANICS.  155 

SOLUTION.—  Reducing  the  miles  to  feet,  and  the  hours  to  seconds,  50 
miles  X  5.280  =  264,000  ft.  H  hours  X  60  X  60  =  5,400  sec.  Applying 
the  rule,  264,000  -H  5,400  =  48|  ft.  per  sec.  Ans. 

467.  If  the  uniform  velocity  (or  the  average  velocity) 
and  the  time  are  given,  and  it  is  required  to  find  the  dis- 
tance which  a  body  having  the  given  velocity  would  travel 
in  the  given  time: 

Rule  73.  —  Multiply  'the  velocity  by  the  time. 

EXAMPLE.—  The  velocity  of  sound  in  still  air  is  1,092  feet  per  second; 
how  many  miles  will  it  travel  in  16  seconds  ? 

SOLUTION.—  Reducing  the  1,092  ft.  to  miles,  1,092  H-  5,280  =  J^. 

o,*oO 

Applying  the  rule,  ^-^r,  X  16  =  3.31  miles,  nearly.     Ans. 


EXAMPLE.  —  The  piston  speed  of  an  engine  is  11  ft.  per  sec.,  how 
many  miles  does  the  piston  travel  in  1  hour  and  15  minutes  ? 

SOLUTION.  —  1    hour   and    15    minutes    reduced    to    seconds  =  4,500 

seconds  =  the  time.     11  feet  reduced  to  miles  =  p-^^  mile  =  velocity 

<J,  ,coU 

in  miles  per  second.    Applying  the  rule,  r         X  4,500  —  9.375  miles.    Ans. 


468.  If  the  distance  through  which  a  body  moves  is 
given,  and  also  its  average  or  uniform  velocity,  and  it  is 
desired  to  know  how  long  it  takes  the  body  to  move  through 
the  given  distance: 

Rule  74.  —  Divide  the  distance,  or  space  passed  over,  by 
the  velocity. 

EXAMPLE.  —  Suppose  that  the  radius  of  the  crank  of  a  steam  engine 
is  15",  and  that  the  shaft  makes  120  revolutions  per  minute,  how  long 
will  it  take  the  crank-pin  to  travel  18,849.6  feet  ? 

SOLUTION.  —  Since  the  radius,  or  distance  from  the  center  of  the 
shaft  to  the  center  of  the  crank-pin,  is  15",  the  diameter  of  the  circle  it 
moves  in  is  15"  X  2  =  30"  =  2.5  ft.  The  circumference  of  this  circle  is 
2.5x3.1416  =  7.854  ft.  7.854x120=942.48  ft.  -  distance  that  the 
crank-pin  travels  in  one  minute  =  velocity  in  feet  per  minute.  Apply- 
ing the  rule,  18,849.6  -H  942.48  =  20  minutes.  Ans. 

EXAMPLE.—  A  point  on  the  rim  of  an  engine  fly-wheel  travels  at  the 
rate  of  150  feet  per  second;  how  long  will  it  take  to  travel  45,000  feet  ? 
SOLUTION.  —  Applying  the  rule, 

45,000  -H  150  =  300  seconds  =  5  minutes.     Ans. 


156  MECHANICS. 

EXAMPLES    FOR    PRACTICE. 

1.  A  locomotive  has  drivers  80'  in  diameter.     If  they  make  293 
revolutions  per  minute,  what  is  the  velocity  of  the  train  in  (a)  feet  per 
second?  (b)  miles  per  hour  ?  .        (  (a)  102.277  ft.  per  sec. 

'((6)    69. 733  mi.  per  hr. 

2.  Assuming  the  velocity  of  steam  as  it  enters  the  cylinder  to  be 
900  feet  per  second,  how  far  could  it  travel,  if  unobstructed,  during 
the  time  the  fly-wheel  of  an  engine  revolved  7  times,  if  the  number  of 
revolutions  per  minute  were  120  ?  Ans.  3,150  ft. 

3.  The  average  speed  of  the  piston  of  an  engine  is  528  feet  per 
minute,  how  long  will  it  take  the  piston  to  travel  4  miles  ? 

Ans.  40  min. 

4.  A  speed  of  40  miles  per  hour  equals  how  many  feet  per  second  ? 

Ans.  58|  ft. 

5.  The  earth  turns  around  once  in  24  hours.     If  the  diameter  be 
taken  as  8,000  miles,   what  is  the  velocity  of  a  point  on  the  earth 
in  miles  per  minute  ?  Ans.  17.45£  mi.  per  min. 

6.  The  stroke  of  an  engine  is  28  inches.     If  the  engine  makes  11,400 
strokes  per  hour,  (a)  what  is  its  speed  in  feet  per  minute  ?    (b)  How 
far  will  this  piston  travel  in  11  minutes  ?        A       (  (a)  443£  ft.  per  min. 

1  (6)  4,876  ft.  8  in. 


FORCE. 

469.  Force  is  that  which  produces,  or  tends  to  pro- 
duce or  destroy,  motion.  Forces  are  called  by  various 
names,  according  to  the  effects  which  they  produce  upon  a 
body,  as  attraction,  repulsion,  cohesion,  adhesion,  accelerating 
force,  retarding  force,  resisting  force,  etc. ,  but  all  are  equiv- 
alent to  a  push  or  pull,  according  to  the  direction  in  which 
they  act  upon  a  body.  That  the  effect  of  a  force  upon  a 
body  may  be  compared  with  another  force,  it  is  necessary 
that  three  conditions  be  fulfilled  in  regard  to  both  bodies. 
They  are  as  follows: 

(1)  The  point  of  application,  or  point  at  which  the  force 
acts  upon  the  body,  must  be  known. 

(2)  The  direction  of  the  force,  or,  what  is  the  same  thing, 
the  straight  line  along  which  the  force  tends  to  move  the  point 
of  application,  must  be  known. 


MECHANICS.  157 

(3)  The  magnitude  or  value  of  the  force,  when  compared 
with  a  given  standard,  must  be  known. 

470.  The  unit  of  magnitude  of  forces  will  always  be 
taken  as  one  pound,  and  all  forces  will  be  spoken  of  as  a 
certain  number  of  pounds. 

In  practice,  force  is  always  regarded  as  a  pressure;  that 
is,  a  force  may  always  be  replaced  by  an  equivalent 
weight.  Thus,  a  force  of  20  Ib.  acting  upon  a  body  is  re- 
garded as  a  pressure  of  20  Ib.  produced  by  a  weight  of  20 
Ib.  The  tendency  of  a  force  is  always  to  produce  motion 
in  the  direction  in  which  it  acts.  The  resistance  may 
be  too  great  for  it  to  cause  motion,  but  it  always  tends  to 
produce  it. 

471.  The    fundamental    principles    of    the    relations 
between  force  and  motion  were  first  stated  by  Sir  Isaac 
Newton.     They    are    called    "Newton's    Three    Laws    of 
Motion,"  and  are  as  follows: 

(1)  All  bodies  continue  in  a  state  of  rest,  or  of  uniform 
motion  in  a  straight  line,  unless  acted  upon  by  some  external 
force  that  compels  a  cJiange. 

(2)  Every  motion  or  change  of  motion  is  proportional  to 
the  acting  force,   and  takes  place   in   the  direction   of  the 
straight  line  along  which  the  force  acts. 

(3)  To  every  action  there  is  always  opposed  an  equal  and 
contrary  reaction. 

472.  In  the  first  law  of  motion  it  is  stated  that  a  body 
once  set  in  motion  by  any  force,  no  matter  how  small,  will 
move    forever    in    a    straight    line,    and    always   with    the 
same   velocity,    unless    acted    upon    by   some   other   force 
which  compels   a   change.     It   is   not   possible   to  actually 
verify   this  law,  on  account  of   the  earth's  attraction  for 
all   bodies,    but,   from    astronomical   observations,    we   are 
certain  that  the  law  is  true.     This  law  is  often  called  the 
law  of  inertia. 


158  MECHANICS. 

473.  The  word  inertia  is  so  abused  that  a  full  under- 
standing of  its  meaning  is  necessary.     Inertia  is  not  a  force, 
although  it  is  often  so  called.     If  a  force  acts  upon  a  body 
and  puts  it  in  motion,  the  effect  of  the  force  is  stored  in  the 
body,  and  a  second  body,  in  stopping  the  first,  will  receive  a 
blow  equal  in  every  respect  to  the  original  force,  assuming 
that  there  has  been  no  resistance  of  any  kind  to  the  motion 
of  the  first  body. 

It  is  dangerous  for  a  person  to  jump  from  a  fast  moving 
train,  for  the  reason  that,  since  his  body  has  the  same  veloc- 
ity as  the  train,  it  has  the  same  force  stored  in  it  that  would 
cause  a  body  of  the  same  weight  to  take  the  same  velocity 
as  the  train,  and  the  effect  of  a  sudden  stoppage  is  the  same 
as  the  effect  of  a  blow  necessary  to  give  the  person  that 
velocity. 

By  "  bracing  "  himself  and  jumping  in  the  same  direction 
that  the  train  is  moving,  and  running,  he  brings  himself 
gradually  to  rest,  and  thus  reduces  the  danger.  If  a  body 
is  at  rest,  it  must  be  acted  upon  by  a  force  in  order  to  be  put 
in  motion,  and,  no  matter  how  great  the  force  may  be,  it 
can  not  be  instantly  put  in  motion. 

The  resistance  thus  offered  to  being  put  in  motion  is  com- 
monly, but  erroneously,  called  the  "Resistance  of  Inertia." 
It  should  be  called  the  Resistance  due  to  Inertia. 

474.  From  the  second  law,  we  see  that  if  two  or  more 
forces  act  upon  a  body,  their  final  effect  upon  the  body  will 
be  in  proportion  to  their  magnitudes,  and  to  the  directions 
in  which  they  act. 

Thus,  if  the  wind  be  blowing  due  West,  with  a  velocity  of 
50  miles  per  hour,  and  a  ball  be  thrown  due  North  with  the 
same  velocity,  or  50  miles  per  hour,  the  wind  will  carry  the 
ball  West  while  the  force  of  the  throw  is  carrying  it  North, 
and  the  combined  effect  will  be  to  cause  it  to  move 
Northwest. 

The  amount  of  departure  from  due  North  will  be  propor- 
tional to.  the  force  of  the  wind,  and  independent  of  the 
velocity  due  to  the  force  of  the  throw. 


MECHANICS. 


159 


475.     In  Fig.  62  a  ball  e  is  supported  in  a  cup,  the  bottom 
of  which  is  attached  to  the  lever  o  in  such  a  manner  that  o 

will  swing  the  bottom 
horizontally  and  allow 
the  ball  to  drop.  An- 
other ball  b  rests  in  a 
horizontal  groove  that 
is  provided  with  a  slit 
in  the  bottom.  A  swing- 
ing arm  is  actuated  by 
the  spring  d  in  such  a 
manner  that,  when 
drawn  back,  as  shown, 
and  then  released,  it  will 
strike  the  lever  o  and 
the  ball  b  at  the  same 
time.  This  gives  b  an 
impulse  in  a  horizontal 
direction,  and  swings  o 
so  as  to  allow  e  to  fall. 

On  trying  the  experi- 
ment, it  is  found  that  b 
follows  a  path  shown  by 
the  curved  dotted  line, 
and  reaches  the  floor  at 
the  same  instant  as  e, 
which  drops  vertically. 
This  shows  that  the 
force  which  gave  the 
first  ball  its  horizontal 
movement  had  no  effect 
on  the  vertical  force 
""-<  which  compelled  both 

balls  to  fall  to  the  floor ;  the  vertical  force  producing  the  same 
effect  as  if  the  horizontal  force  had  not  acted.  The  second 
law  may  also  be  stated  as  follows:  A  force  has  the  same  effect 
in  producing  motion,  whether  it  acts  upon  a  body  at  rest  or  tn 
motion,  and  -whether  it  acts  alone  or  with  other  forces. 


\ 


160  MECHANICS. 

476.  The  third  law  states  that  action  and  reaction  are 
equal  and  opposite.     A  man  can  not  lift  himself  by  his  boot- 
straps, for  the  reason  that  he  presses  downwards  with  the 
same  force  that  he  pulls  upwards;  the  downward  reaction 
equals  the  upward  action,  and  is  opposite  to  it. 

In  springing  from  a  boat  we  must  exercise  caution  or  the 
reaction  will  drive  the  boat  from  the  shore.  When  we  jump 
from  the  ground,  we  tend  to  push  the  earth  from  us,  while 
the  earth  reacts  and  pushes  us  from  it. 

EXAMPLE. — Two  men  pull  on  a  rope  in  opposite  directions,  each  ex- 
erting a  force  of  100  pounds;  what  is  the  force  which  the  rope  resists  ? 

SOLUTION.— Imagine  the  rope  to  be  fastened  to  a  tree,  and  one  man 
to  pull  with  a  force  of  100  pounds.  The  rope  evidently  resists  100 
pounds.  According  to  Newton's  third  law,  the  reaction  of  the  tree 
is  also  100  pounds.  Now,  suppose  the  rope  to  be  slackened,  but  that 
one  end  is  still  fastened  to  the  tree,  and  the  second  man  to  take  hold 
of  the  rope  near  the  tree,  and  pull  with  a  force  of  100  pounds,  the  first 
man  pulling  as  before.  The  resistance  of  the  rope  is  100  pounds,  as 
before,  since  the  second  man  merely  takes  the  place  of  the  tree.  He 
is  obliged  to  exert  a  force  of  100  pounds  to  keep  the  rope  from  slip- 
ping through  his  fingers.  If  the  rope  be  passed  around  the  tree,  and 
each  man  pulls  an  end  with  a  force  of  100  pounds  in  the  same  and 
parallel  directions,  the  stress  in  the  rope  is  100  pounds,  as  before,  but 
the  tree  must  resist  the  pull  of  both  men,  or  200  pounds. 

477,  A  force  may  be  represented  by  a  line;  thus,  in 
Fig.  63,  let  A  be  tf\&  point  of  application  of 

the  force ;  let  the  length  of  the  line  A  D  A  B 

represent  its  magnitude,  and  let  the  arrow-  FIG.  63. 

head  indicate  the  direction  in  which  the  force  acts,  then  the 
line  A  B  fulfils  the  three  conditions  (see  Art.  469),  and  the 
force  is  fully  represented. 


CENTER  OF  GRAVITY. 

478.  The  center  of  gravity  of  a  body  is  that  point  at 
which  the  body  may  be  balanced,  or  it  is  the  point  at  whicli 
the  'Mliole  weight  of  a  body  may  be  considered  as  concentrated. 

This  point  is  not  always  in  the  body;  in  the  case  of  a 
horseshoe  or  a  ring  it  lies  outside  of  the  substance  of,  but 
within  the  space  enclosed  by,  the  body. 


MECHANICS. 


161 


In  a  moving  body,  the  line  described  by  its  center  of 
gravity  is  always  taken 
as  the  path  of  the  body. 
In  finding  the  distance 
that  a  body  has  moved, 
the  distance  that  the 
center  of  gravity  has 
moved  is  taken. 

The  definition  of  the 
center  of  gravity  of  a  FIG- 64- 

body   may   be   applied   to  a  system  of  bodies  if   they  are 
considered  as  being  connected  at  their  centers  of  gravity. 

If  iv  and  W,  Fig.  64,  be  two  bodies  of  known  weight,  their 
center  of  gravity  will  be  at  C.  The  point  C  may  be  readily 
determined  as  follows: 

Rule  75. —  The  distance  of  the  common  center  of  gravity, 
from  the  center  of  gravity  of  the  large  weight,  is  equal  to  the 
weight  of  the  smaller  body  multiplied  by  the  distance  between 
the  centers  of  gravity  of  the  tivo  bodies,  and  this  product 
divided  by  the  sum  of  the  weights  of  the  two  bodies. 

EXAMPLE.— In  Fig.  64,  a/  =  10  pounds,  W='&  pounds,  and  the  dis- 
tance between  their  centers  of  gravity  is  36  inches  where  is  the  center 
of  gravity  of  both  bodies  sit- 
uated ? 

SOLUTION.— Applying  the 
rule,  10  x  36  =  360.  10  +  30= 
40.  360  -T-  40  =  9"  =  distance 
of  center  of  gravity  from  cen- 
ter of  large  weight. 

479.  It  is  now  very 
easy  to  extend  this  prin- 
ciple, to  find  the  center 
of  gravity  of  any  num- 
ber of  bodies  when  their 
weights  and  the  dis- 
tances apart  of  their  cen- 
ters of  gravity  are  known,  by  the  following  rule: 

Rule  76. — Find  the  center  of  gravity  of  two  of  the  bodies 
as  W^  and  W^,  in  Fig.  65.  Assume  that  the  weight  of  both 


162  MECHANICS. 

bodies  is  concentrated  at  £*„  and  find  the  center  of  gravity  of 
this  combined  weight  at  C^  and  the  weight  of  Wj  let  it  be 
at  C9;  then  find  the  center  of  gravity  of  the  combined  weights 
of  W^,  W»  W9  (concentrated  at  Q,  and  W^;  let  it  be  at 
C ;  then  C  will  be  the  center  of  gravity  of  the  four  bodies. 

480.  To  find   the  center  of  gravity  of  any  parallel- 

ogram : 

Rule  77. — Draw  the  two  diago- 
nals, Fig.  66,  and  their  point  of 
intersection  C  will  be  the  center  of 
gravity. 

481.  To  find  the  center  of  gravity   of  a  triangle,  as 
ABC,  Fig.  67  : 

Rule  78. — From  any  vertex, 
as  A,  draw  a  line  to  the  middle 
point  D  of  the  opposite  side  B  C. 
From  one  of  the  other  vertexes, 
as  C,  draw  a  line  to  F,  the  mid- 
dle point  of  the  opposite  side  A  B  ; 
the  point  of  intersection  O  of  these  two  lines  is  the  center 
of  gravity. 

It  is  also  true  that  the  distance  D  O  —\  D  A,  and  that 
F  O  =  $  F  C,  and  the  center  of  gravity  could  have  been 
found  by  drawing  from  any  vertex  a  line  to  the  middle 
point  of  the  opposite  side,  and  measuring  back  from  that 
side  $  of  the  length  of  the  line. 

The  center  of  gravity  of  any  regular  plane  figure  is 
the  same  as  the  center  of  the  inscribed  or  circumscribed 
circle. 

482.  To  find  the  center  of  gravity  of  any  irregular 
plane  figure,  but  of  uniform  thickness  throughout,  divide 
one  of  the  parallel  surfaces  into  triangles,  parallelograms, 
circles,  ellipses,  etc.,  according  to  the  shape  of  the  figure; 
find   the   area   and   center  of  gravity  of    each   part   sepa- 
rately, and  combine  the  centers  of    gravity  thus  found  in 
the   same  manner    as    in    rule  76 ;   in  this   case,  however, 


MECHANICS. 


163 


dealing  with   the  area  of  each  part  instead  of  its  weight. 
See  Fig.  68. 


EXAMPLE.— Suppose  that  the  two  balls  shown  in  Fig.  64  were  5*  and 
10"  in  diameter,  and  weighed  10  Ib.  and  80  lb.,  respectively.  If  the  dis- 
tance between  their  centers  were  40",  and  they  were  connected  by  a  steel 
rod  1*  in  diameter,  where  would  the  center  of  gravity  be,  taking  the 
weight  of  a  cubic  inch  of  steel  as  .283  lb.  ? 

SOLUTION.— The  length  of  the  rod  =  40  —  y  —  -5-  =  32i"-  Its  volume 

is  1>  X  .7854  X  32*  =  25.53  cu.  in.  25.53  X  .283  =  7.22  lb.  The  rod 
being  straight,  its  center  of  gravity  is  in  the  middle  at  a  distance  of 

-£-  +  -g  =  18f"  from  the  center  of  the  smaller  weight,  and  '-^—  +  -g-  = 

21J*  from  the  center  of  the  larger  weight.  Now,  considering  the  weight 
of  the  rod  to  be  concentrated  at  its  center  of  gravity,  we  have  three 
weights  of  10,  7.22,  and  80  lb.,  all  in  a  straight  line,  and  the  distances 
between  them  given  to  find  the  center  of  gravity,  or  balancing  point, 
of  the  combination,  by  rule  76.  We  will  first  find  the  center  of  gravity 
of  the  two  smaller  weights  by  rule  75,  as  follows:  7.22  X  18}  =  135.38. 
10  +  7.22  =  17.22.  135.38  -*•  17.22  =  7.86"  =  distance  from  the  center  of 
the  10-lb.  weight.  Considering  both  of  the  smaller  weights  to  be  con- 
centrated at  this  point,  we  find  the  center  of  gravity  of  this  combined 
weight  and  the  large  weight  as  follows.  40  —  7.86  =  32.14"  =  distance 


164 


MECHANICS. 


between  the  center  of  gravity  of  the  two  small  weights  and  the  center 
of  gravity  of  the  80-lb.  weight.  Applying  rule  75,  17.22  X  32.14  = 
553.45. 

17.22  +  80  =  97.22.    553.45  H-  97.22  =  5.693"  =  distance  from  the  center 
of  the  80-lb.  weight.     Ans. 

483.     Center  of  Gravity  of  a  Solid. — In  a  body  free 
to  move,  the  center  of  gravity  will  lie  in  a  vertical  plumb 


SlG,  69.  \ 

line  drawn  through  the  point  of  support.  Therefore,  to 
find  the  position  of  the  center  of  gravity  of  an  irregular 
solid,  as  the  crank,  Fig.  69,  suspend  it  at  some  point,  as  B, 
so  that  it  will  move  freely.  Drop  a  plumb  line  from  the 
point  of  suspension,  and  mark  its  direction.  Suspend  the 
body  at  another  point,  as  A,  and  repeat  the  process.  The 
intersection  C  of  the  two  lines  will  be  directly  over  the 
center  of  gravity. 

Since  the  center  of  gravity  depends  wholly  upon  the  shape 
and  weight  of  a  body,  it  may  be  without  the  body,  as  in  the 
case  of  a  circular  ring  whose  center  of  gravity  is  the  same 
as  the  center  of  the  circumference  of  the  ring. 


EXAMPLES   FOR  PRACTICE. 

1.  A  spherical  shell  has  a  wrought-iron  handle  attached  to  it.  The 
shell  is  10"  in  diameter,  and  weighs  20  Ib.  The  handle  is  H"  in  diameter, 
and  the  distance  from  the  center  of  the  shell  to  the  end  of  the  handle 
is  4  ft.  Where  is  the  center  of  gravity  ?  Take  the  weight  of  a  cubic 
inch  of  wrought  iron  as  .278  Ib.  Ans.  13.612"  from  center  of  shell. 


MECHANICS.  165 

2.  The  distance  between  the  centers  of  two  bodies  is  51".     The 
weights  of  the  bodies  being  20  and  73  lb.,  where  is  the  center  of  gravity  ? 

Ans.  10.968"  from  the  center  of  large  weight. 

3.  A  hollow  engine  piston  weighs  275  lb.,  and  is  3£"  thick.     Assum- 
ing the  piston  rod  to  be  straight  throughout  its  entire  length,  and  to 
weigh  140  lb.,  at  what  point  will  the  piston  and  rod  balance  if  the 
length  of  the  rod  is  73"  from  the  face  of  the  piston  ?    Consider  the 
weight  of  the  piston  to  be  concentrated  at  its  center. 

Ans.  11.15",  nearly,  from  face  of  piston. 

4.  Weights  of  5,  9,  and  12  lb.  lie  in  one  straight  line  in  the  order 
named.     Distance  from  the  5-lb.  weight  to  the  9-lb.  weight  is  22",  and 
from  the  9-lb.  weight  to  the  12-lb.  weight  is  18".     Where  is  the  center 
of  gravity?  Ans.  13.923"  from  12-lb.  weight. 

SIMPLE    MACHINES. 

484.     A  lever  is  a  bar  capable  of  being  turned  about  a 
pivot,  or  point,  as  in  Figs.  70,  71,  and  72. 


lw\ 

FIG.  70.  FIG.  71.  FIG.  73. 

The  object  W  to  be  lifted  is  called  the  weight ;  the 
force  employed  P  is  called  the  power ;  and  the  point,  or 
pivot,  F  is  called  the  fulcrum. 

485.  That  part  of  the  lever  between  the  fulcrum  and 
the  weight,  or  F  b,  is  called  the  weight  arm,  and  the  part 
between  the  fulcrum  and  the  power,  or  /'"  c,  is  called  the 
power  arm. 

In  order  that  the  lever  shall  be  in  equilibrium  (balance), 
the  power  multiplied  by  t tie  power  arm  must  equal  tlie  weight 
multiplied  by  the  weight  arm;  that  is,  Px  F c  must  equal 
WxFb. 

486.  If  F  be  taken  as  the  center  of  a  circle,  and  arcs  be 
described  through  b  and  c,  it  will  be  seen  that  if  the  weight 
arm  be  moved  through  a  certain  angle,  the  power  arm  will 
move  through  the  same  angle.     Since,  in  the  same  or  equal 
angles,  the  lengths  of  the  arcs  are  proportional  to  the  radii 


100  MECHANICS. 

with  which  they  were  described,  it  is  seen  that  the  power 
arm  is  proportional  to  the  distance  through  which  the 
power  moves,  and  the  weight  arm  is  proportional  to  the 
distance  through  which  the  weight  moves. 

Hence,  instead  of  writing  Px  Fc=  JTx  F  b,  we  might 
have  written  it  PX  distance  through  which  P  moves  = 
W  X  distance  through  which  W  moves.  This  is  the  general 
law  of  all  machines,  and  can  be  applied  to  any  mechanism 
from  the  simple  lever  up  to  the  most  complicated  arrange- 
ment. Stated  in  the  form  of  a  rule,  it  is  as  follows: 

Rule  79. —  The  power  multiplied  by  tlie  distance  through 
which  it  moves  equals  tlie  weiglit  multiplied  by  tJie  distance 
through  which  it  moves. 

487.  In  the  above  rule,  it  will  be  noticed  that  there  are 
four  requirements  necessary  for  a  complete  knowledge  of 
the  lever,  viz. :  the  power  (or  force),  the  weight,  the  power 
arm  (or  distance  through  which  the  power  moves),  and  the 
weight  arm  (or  distance  through  which  the  weight  moves). 
If  any  three  are  given,  the  fourth  may  be  found  by  letting 
x  represent  the  requirement  which  is  to  be  found,  and  mul- 
tiplying the  power  by  the  power  arm  and  the  weight  by  the 
weight  arm  ;  then,  dividing  the  product  of  the  two  known 
numbers    by    the    number    by   which  x  is  multiplied  ;  the 
result  will  be  the  requirement  which  was  to  be  found. 

EXAMPLE. — If  the  weight  arm  of  a  lever  is  6  in.  long,  and  the  power 
arm  is  4  ft.  long,  how  great  a  weight  can  be  raised  by  a  force  of  20  Ib. 
at  the  end  of  the  power  arm  ? 

SOLUTION. — In  this  example,  the  weight  is  unknown;  hence,  repre- 
senting it  by  .r,  we  have,  after  reducing  the  4  ft.  to  inches,  20  X  48  = 
960  =  power  multiplied  by  the  power  arm,  and  x  x  6  =  weight  multi- 
plied by  the  weight  arm.  Dividing  the  960  by  6,  the  result  is  160  Ib., 
the  weight.  Ans. 

488.  If  the  distance  through  which  the  power  or  weight 
moved  had  been  given  instead  of  the  power  arm  or  weight 
arm,,  and  it  were  required  to  find  the  power  or  weight,  the 
process  would  have  been  exactly  the  same,  using  the  given 
distance  instead  of  the  power  arm  or  weight  arm. 


MECHANICS. 


167 


EXAMPLE. — If,  in  the  above  example,  the  weight  had  moved  2^*, 
how  far  would  the  power  have  moved  ? 

SOLUTION.— In  this  example,  the  distance  through  which  the  power 
moves  is  required.  Let  .r  represent  the  distance.  Then,  20  x  x  —  dis- 
tance multiplied  by  power,  and  2^  x  16')  =  400  =  distance  multiplied  by 

400 
the  weight.     Hence,  ,t  =  -^-  =  20  in.  =1.-  distance   through   which   the 

power  arm  moves. 

The  ratio  between  the  weights  and  the  power  is  160  -*-  20  =  8.  The 
ratio  between  the  distance  through  which  the  weight  moves  and  the 
distance  through  which  the  power  moves  is  2^  -*-  20  =  £.  This  shows 
that  while  a  force  of  1  Ib.  can  raise  a  weight  of  8  lb.,  the  1-lb.  weight 
must  move  through  8  times  the  distance  that  the  8-lb.  weight  does.  It 
will  also  be  noticed  that  the  ratio  of  the  lengths  of  the  two  arms  of  the 
lever  is  also  8,  since  48  -*-  6  =  8. 

489.  The  law  which  governs  the  straight  lever  also 
governs  the  bent  lever,  but  care  must  be  taken  to  determine 
the  true  lengths  of  the  lever  arms,  which  are  in  every  case 
the  perpendicular  distances  from  the  fulcrum  to  the  line  of 
direction  of  the  weight  or  power. 
c  F  b 


FIG.  75.  ["  \  FIG  7tj. 

Thus,  in  Figs.  73,  74,  75,  and  7G,  Fc  in  each  case  represents 
the  power  arm,  and  F  b  the  weight  arm. 

49O.     A  compound  lever  is  a  series  of  single  levers 
arranged  in  such  a  manner  that  when  a  power  is  applied  to 


168  MECHANICS. 

the  first  it  is  communicated  to  the  second,  and  from  that  to 
the  third,  and  so  on. 

Fig.  77  shows  a  compound  lever.     It  will  be  seen  that 
when  a  power  is  applied  to  the  first  lever  at  Pit  will  be 


ff. 

6'\ -3Qg — 


communicated  to  the  second  lever  at  P,  from  this  to  the 
third  lever  at  P,  and  thus  raise  the  weight  W. 

The  weight  which  the  power  of  the  first  lever  could  raise 
acts  as  the  power  of  the  second,  and  the  weight  which  this 
could  raise  through  the  second  lever  acts  as  the  power  of 
the  third  lever,  and  so  on,  no  matter  how  many  single  levers 
make  up  the  compound  lever. 

In  this  case,  as  in  every  other,  the  power  multiplied  by 
the  distance  through  which  it  moves  equals  the  weight 
multiplied  by  the  distance  through  which  it  moves. 

Hence,  if  we  move  the  ^Pend  of  the  lever,  say  4  inches, 
and  the  end  carrying  the  weight  W  moves  \  of  an  inch,  we 
know  that  the  ratio  between  P  and  W  is  the  same  as  the 
ratio  between  \  and  4;  that  is,  1  to  20,  and,  hence,  that 
10  pounds  at  P  would  balance  200  pounds  at  IV,  without 
measuring  the  lengths  of  the  different  lever  arms.  If  the 
lengths  of  the  lever  arms  are  known,  the  ratio  between 
P  and  W  may  be  readily  obtained  from  the  following  rule: 

Rule  8O. —  The  continued  product  of  the  power  and  each 
power  arm  equals  the  continued  product  of  the  weight  and 
each  weight  arm. 

EXAMPLE.— If,  in  Fig.  77,  P  F—  24  inches,  18  inches,  and  30  inches, 
respectively,  and  W  F=  6  inches,  6  inches,  and  18  inches,  respectively, 
how  great  a  force  at  P  would  it  require  to  raise  1,000  pounds  at  W? 
What  is  the  ratio  between  ffand  PI 

SOLUTION.— Let  x  represent  the  power;  then,  .r  x  24  x  18  X  30  = 
12,960 .r  =  continued  product  of  the  power  and  each  power  arm.  l.OOOx 


MECHANICS. 


169 


6  X  6  X  18  =  648,000  =  continued  product  of  the  weight  and  each  weight 
arm,  and,  since  12,960  .r=  648,000, 
648,000 

•*" =  10  Qftn  =  50  lb-  =  tne  Power-     Ans- 
i*,you 

1,000  -5-  50  =  20  =  ratio  between  Wand  P. 
491.     The  wheel  and  axle  consists  of  two  cylinders  of 

different    diameters,   rigidly  connected,   so    that   they    turn 
together  about  a  common  axis,  as  in   Fig.  78.      Then,  as 


before,  P  X  distance  through  which  it  moves  =  W  X  dis- 
tance through  which  it  moves;  and,  since  these  distances 
are  proportional  to  the  radii  of  the  power  cylinder  and 
weight  cylinder,  P  X  F  c  =  } V  x  F  b. 

It  is  not  necessary  that  an  entire  wheel  be  used ;  an  arm, 
projection,  radius,  or  anything  which  the  power  causes  to 
revolve  in  a  circle,  may  be  con- 
sidered as  the  wheel.  Conse- 
quently, if  it  is  desired  to  hoist 
a  weight  with  a  windlass,  Fig. 
79,  the  power  is  applied  to  the 
handle  of  the  crank,  and  the  dis- 
tance between  the  center  line  of 
the  crank  handle  and  the  axis  FIG.  79. 

of  the  drum  corresponds  to  the  radius  of  the  wheel. 

EXAMPLE. — If  the  distance  between  the  center  line  of  the  handle  and 
the  axis  of  the  drum,  in  Fig.  79,  is  18  inches  and  the  diameter  of  the 
drum  is  6  inches,  what  force  will  be  required  at  P  to  raise  a  load  of  300 
pounds  ? 

SOLUTION.—/*  x  (18  X  2)  =  300  x  6,  or  P  =  50.     Ans. 


170 


MECHANICS. 


EXAMPLES   FOR   PRACTICE. 

1.  The  lever  of  a  safety  valve  is  of  the  form  shown  in  Fig.  70,  where 
the  force  is  applied  at  a  point  between  the  fulcrum  and  the  weight 
lifted.  If  the  distance  from  the  fulcrum  to  the  valve  is  5£",  and  from 
the  fulcrum  to  the  weight  is  42",  what  total  force  is  necessary  to  raise 
the  valve,  the  weight  being  78  lb.,  and  the  weight  of  valve  and  lever 
being  neglected  ?  Ans.  595.64  lb. 

•  2.  If,  in  Fig.  77,  P  F-  10",  12",  14",  and  16"  respectively,  and  W F= 
2",  3",  4",  and  5",  respectively,  (a)  how  great  a  weight  can  a  force  of  20  lb. 
raise?  (b)  What  is  the  ratio  between  Wand  /*?  (c)  If  Amoves  4", 
how  far  will  W  move  ?  t  (a)  4,480  lb. 

Ans.  ]  (£)  224. 

<  (c)  5y. 

3.  A  windlass  is  used  to  hoist  a  weight.  If  the  diameter  of  the  drum 
on  which  the  rope  winds  is  4",  and  the  distance  from  the  center  of  the 
handle  to  the  axis  of  the  drum  is  14",  how  great  a  weight  can  a  force  of 
32  lb.  applied  to  the  handle  raise  ?  Ans.  224  lb. 


PULLEYS. 

492.  Pulleys  for  the  transmission  of  power  by  belts 
may  be  divided  into  two  principal  classes:  1.  The  solid 
pulley  shown  in  Fig.  80,  in  which  the  hub,  arms,  and  rim  are 


one  entire  casting.     2.   The  split  pulley  shown  in  Fig.  81, 
which  is  cast  in  halves. 


MECHANICS.  171 

The  latter  style  of  pulley  is  more  readily  placed  upon  and 
removed  from  the  shaft  than  the  solid  pulley.  Pulleys  are 
generally  cast  in  halves  or  parts  when  they  are  more  than 
six  feet  in  diameter;  this  is  done  on  account  of  shrinkage 
strain  in  large  pulley  castings,  which  renders  them  liable  to 
crack  as  a  result  of  unequal  cooling  of  the  metal. 

493.  Crowning. — In  Fig.  82  is  shown  a  section  of  the 
rim  of  a  pulley  that  has  crowning,  or,  in  other  words,  whose 
diameter  is  larger  at  the  center  of  the 

face  than  at  its  edges.  This  is  done  to 
prevent  the  belt  from  running  off  the 
pulley.  The  amount  of  crowning 
given  to  pulleys  varies  from  fa  to  \  an 
inch  per  foot  of  width  of  the  pulley  face.  FlG-  82- 

494.  Balanced   Pulleys. — All   pulleys  which   rotate 
at  high  speeds  should  be  balanced.     If  they  are  not,  the 
centrifugal  force  which  is  generated  by  the  pulley's  rotation 
is  greater  on  one  side  than  on  the  other,  and  it  will  cause 
the  pulley  shaft  to  vibrate  and  shake.      Pulleys  should  run 
true,  so  that  the  strain,  or  tension,  of  the  belt  is  equal  at  all 
parts  of  the  revolution,  thus  making  the  transmitting  power 
equal.     The  smoother  the  surface  of  a  pulley,  the  greater 
is  its  driving  power. 

The  transmitting  power  of  a  pulley  can  be  increased  by 
covering  the  face  of  the  pulley  with  a  leather  or  rubber 
band;  this  increases  the  driving  power  about  one-quarter. 

495.  The  pulley  that  imparts  motion  to  the  belt  is  called 
the  driver ;  that   which  receives  the  motion  is  called   the 
driven. 

The  revolutions  of  any  two  pulleys  over  which  a  belt  is 
run  vary  in  an  inverse  proportion  to  their  diameters;  con- 
sequently, if  a  pulley  of  20  inches  in  diameter  is  driven  by 
one  of  10  inches  in  diameter,  the  20-inch  pulley  will  make 
one  revolution  while  the  10-inch  pulley  makes  two  revolu- 
tions, or  they  are  in  the  ratio  of  2  to  1.  From  this  the 
following  formulas  have  been  deduced: 


172  MECHANICS. 

Let  D  =  diameter  of  the  driver; 
d=  diameter  of  the  driven; 
N=  number  of  revolutions  of  the  driver; 
n  —  number  of  revolutions  of  the  driven. 

NOTE.  —  The  words  revolutions  per  minute  are  frequently  abbrevi- 
ated to  R.  P.  M. 

496.  To  find  the  diameter  of  the  driving  pulley  when  the 
diameter  of  the  driven  pulley  and  the  number  of  revolutions 
per  minute  of  each  is  given: 

Rule  81.  —  The  diameter  of  the  driving  pulley  equals 
the  product  of  the  diameter  and  number  of  revolutions 
of  the  driven  pulley  divided  by  the  number  of  revolu- 
tions of  the  driving  pulley. 

That  is,  D  =  j£. 

EXAMPLE.  —  The  driving  pulley  makes  100  revolutions  per  minute, 
the  driven  pulley  makes  75  revolutions  per  minute,  and  is  18  inches 
in  diameter  ;  what  is  the  diameter  of  the  driving  pulley  ? 

SOLUTION.  —  Formula,  D  =  -=-. 

1  Q  vx  f7P| 

Substituting,  we  have  D  =  —  jgg  —  =  13J  in.     Ans. 

497.  The   diameter   and    number  of    revolutions  per 
minute  of  the  driving  pulley  being  given,  to  find  the  diam- 
eter of  the  driven  pulley,  which  must  make  a  given  number 
of  revolutions  per  minute: 

Rule  82.  —  The  diameter  of  the  driven  pulley  equals 
the  prodtict  of  the  diameter  and  number  of  revolutions 
of  the  driving  pulley  divided  by  the  number  of  revolu- 
tions of  the  driven  pulley. 


Thatis,  ,, 

n 

EXAMPLE.  —  The  diameter  of  the  driver  is  13£  inches,  and  makes 
100  revolutions  per  minute  ;  what  must  be  the  diameter  of  the  driven 
to  make  75  revolutions  per  minute  ? 

SOLUTION.  —  Formula,  d=  -  . 

n 

Substituting,  we  have  d=  13^  *  10°  =  13  in.     Ans. 
<o 


MECHANICS.  173 

498.  To  find  the  number  of  revolutions  per  minute  of  the 
driven  pulley,  its  diameter  and  the  diameter  and  number  of 
revolutions  per  minute  of  the  driving  pulley  being  given: 

Rule  83.  —  The  number  of  revolutions  of  the  driven  pulley 
equals  the  product  of  the  diameter  and  the  number  of  revolu- 
tions of  the  driver  divided  by  the  diameter  of  the  driven  pulley. 

That  is, 


EXAMPLE.  —  The  driver  is  Ifrt  inches  in  diameter,  and  makes 
100  revolutions  per  minute;  how  many  revolutions  will  the  driven 
make  in  one  minute,  if  it  is  18  inches  in  diameter  ? 

D  N 

SOLUTION..—  Formula,  n  =  —  -T-. 

a 


Substituting,  we  have  n  =  J"2  ~    —  =  75  R.  P.  M.     Ans. 

499.  To  find  the  number  of  revolutions  per  minute  of 
the  driving  pulley,  its  diameter  and  the  diameter  and  num- 
ber of  revolutions  per  minute  of   the  driven  pulley  being 
given : 

Rule  84. —  The  number  of  revolutions  of  the  driving 
pulley  equals  the  product  of  the  diameter  and  number  of 
revolutions  of  the  driven  pulley  divided  by  the  diameter 
of  the  driving  pulley. 

That  is,  N=*g. 

EXAMPLE. — The  driven  pulley  is  18  inches  in  diameter,  and  makes 
75  revolutions  per  minute;  how  many  revolutions  will  the  driver  make 
in  one  minute,  if  it  is  13£  inches  in  diameter  ? 

SOLUTION.  —Formula, 

Substituting,  we  have  N=  '"  ^  '"  =  100  R.  P.  M.    Ans. 

500.  Wheel-work. — When  there  is  a  combination  of 
wheels  and  axles,  as  in  Fig.  83,  it  is  called  a  train.     The 
wheel  to  which  the  power  is  applied,  that  is,  the  one  which 
imparts  the  motion,  is  called   the  driver;  that  which  re- 
ceives it,  the  driven,  or   follower,  and   the  small  wheel 
upon  the  axle,  the  pinion. 

501.  It  will  be  seen  that  the  wheel  and  axle  bears  the 
same    relation  to  the  train  that  the  simple  lever  does   to 


174  MECHANICS. 

the  compound  lever.     Letting  Z?,,  D^  Da,  etc.,  represent 


PIG.  83. 


the  diameters  of  the  different  drivers,  and  dlt  d^  d^  etc., 
the  diameters  of  the  different  pinions,  we  have  the  following: 

Rule  85. —  The  continued  product  of  the  power  and  the 
radii  of  the  drivers  equals  the  continued  product  of  the  weight, 
the  radius  of  the  drum  that  moves  the  weight,  and  the  radii 
of  the  pinions. 

That  is,  P  X  D,  X  D^  X  D»  etc.,=  W  X  ^  X  <  X  </„  etc. 

EXAMPLE.— If  the  radius  of  the  pulley  A  is  20  inches;  of  C,  15 
inches,  and  of  £,  24  inches,  and  the  radius  of  the  drum  J\  is  4  inches; 
of  the  pinion  D,  5  inches,  and  of  the  pinion  B,  4  inches,  how  great  a 
weight  will  a  force  of  1  pound  at  P  raise  ? 

SOLUTION.— Formula,  Px£>iX£>tX&t=  WX  </i  X  <*«  X  <£>• 
Substituting,  we  have  1  X  20  X  15  X  24  =  IV  x  4  X  5  X  4,  or 

Ans. 


MECHANICS.  175 

Hence,  also,  if  W  were  raised  1  inch,  P  would  fall 
90  inchesj  or  P  would  have  to  move  through  90  inches  to 
raise  W  through  1  inch. 

502.  It  is  now  clear  that  another  great  law  has  made 
itself  manifest,  and  that  is  that  whenever  there  is  a  gain  in 
power  without  a  corresponding  increase  in  the  initial  force, 
there  is  a  loss  in  speed ;  this  is  true  of  any  machine. 

In  the  last  example  if  P  were  to  move  the  entire  90  inches 
in  one  second,  W  would  move  only  1  inch  in  one  secon'd. 

Instead  of  using  the  diameter  or  radius  of  a  gear,  the 
number  of  teeth  may  be  used  when  computing  the  weight 
which  can  be  raised,  or  the  velocity,  as  in  the  last  example. 

EXAMPLE.— The  radius  of  the  pulley  A,  Fig.  83,  is  40",  and  that  of 
fis  12".  The  number  of  teeth  in  B  is  9;  in  C,  27;  in  /?,  12,  and  in  E, 
36.  If  the  weight  to  be  lifted  is  1,800  lb.,  how  great  a  force  at  P  is  it 
necessary  to  apply  to  the  belt  ? 

SOLUTION. — Let  /"represent  the  force  (power);  then,  by  the  rule, 
85,  P  X  40  X  27  x  36  =  1,800  X  12  X  9  x  12,  or  P  x  38,880  =  2,332,800. 

O  QQO  xi  h  I 

Hence,  P  =     'gg  g'gQ     =  60  lb.  =  force  necessary  to  apply  to  the  belt. 

Ans. 
GEAR-WHEELS. 

503.  A  wheel  that  is  provided  with  teeth  to  mesh  with 
similar     teeth     upon     another 

wheel  is  called  a  gear-wheel, 
or  gear.  In  Fig.  84  is  shown 
a  spur  gear.  On  spur  gears 
the  teeth  are  always  parallel  to 
the  axis  of  the  wheel  or  to  its 
shaft. 

504.  In  Fig.  85  is  shown  a 
pair  of  bevel  gears  in  mesh, 
of  which  one  is  smaller  than  the 
other. 

505.  When  both  are  of  the 
same  diameter  they  are  called 

FIG.  IM. 

miter  gears. 

In  Fig.  80  is  shown  a  pair  of  miter  gears  in  mesh.     It 


176 


MECHANICS. 


obvious  that  the  angle  which  the  teeth  of  these  gears  make 
with  the  axis  of  the  shaft  must  be  45°. 

5O6.  In  Fig.  87  is  shown  a  revolving  screw,  or  worm,  as  it  is 

called,  in  gear;  it  is  used 
to  transmit  motion  from 
one  shaft  to  another  at 
right  angles  to  it. 

As  the  worm  is  noth- 
ing else  than  a  screw, 
each  revolution  given 
to  the  worm  will  rotate 
the  worm-wheel  a  dis- 
tance equal  to  its  pitch ; 

FIG.  85.  consequently,    if    there 

are  40  teeth  in  the  worm-wheel,  a  single-threaded  worm  will 


FIG.  86. 
have  to  make  40  revolutions  in  order  to  turn  the  wheel  once. 


MECHANICS. 


177 


5O7.  In  Fig.  88  is  shown  a  section  of  a  rack  and  pinion, 
both  having  epicycloidal  teeth.  The  arc  C  C  represents 
part  of  the  pitch  circle  ;  it  is  on  the  pitch  circle  that  all 
the  teeth  are  laid  out.  The  diameter  of  a  gear  or  worm- 
wheel  is  always  taken  as  the  diameter  of  this  circle,  unless 
otherwise  specially  stated  as  "diameter  over  all,"  or 
"diameter  at  the  root,"  etc. 

The  pitch  of  the  teeth  of 
the  gear-wheel  is  the  distance 
from  the  edge  of  one  tooth  to 
the  corresponding  edge  of  the 
following  tooth  measured  on 
the  pitch  circle;  it  is  marked 
pitch  in  the  figure. 

The  length  of  the  tooth  of  a 
gear-wheel  is  .7  of  its  pitch,  .4 
of  it,  called  the  root,  being  be- 
low or  within  the  pitch  circle, 
and  .3  of  it,  called  the  ad- 
dendum, being  above  or  with- 
out the  pitch  circle.  Thus,  if 
the  pitch  of  the  teeth  of  a  gear-  FIG- 87- 

wheel    is   2    inches,  the  length  of   a  tooth  below  the  pitch 
circle  is  2  X  .4=  .8  of  an  inch;  and  its  length  above  the 


FIG* 


pitch   circle   is  2  X.  3  =  .6   of  an   inch.     Consequently,  we 


178  MECHANICS. 

have  only  to  multiply  the  pitch  by  .4  to  obtain  the  length  of 
the  teeth  below  the  pitch  circle,  and  by  .3  to  obtain  the 
length  of  the  teeth  above  the  pitch  circle.  The  thickness  of 
the  teeth  of  a  cast  gear-wheel  equals  .48  X  /*,  that  is,  .48 
of  the  pitch ;  therefore,  the  thickness  of  the  above  teeth  is 
.48  X  2,  or  .96  of  an  inch. 

A  rack  may  be  considered  as  a  gear-wheel  rolled  out  so  as 
to  make  the  pitch  circle  a  straight  line,  as  C'  C".  The  teeth 
of  racks  are  proportioned  by  the  same  rules  as  those  of 
gear-wheels. 

5O8.  For  the  purpose  of  calculating  the  pitch  diameter, 
number  of  teeth,  etc.,  of  the  gear-wheels,  we  have  the 
following  rules: 

Let  P  =  pitch ; 

T=  number  of  teeth; 
D  =  pitch  diameter  of  the  wheel. 

To  find  the  pitch  diameter  of  a  gear-wheel  in  inches,  when 
the  pitch  and  number  of  teeth  are  given: 

Rule  86. —  The  pitch  diameter  equals  the  product  of  the 
pitch  and  number  of  teeth  divided  by  3. 1416. 

Thatis>  D= 


EXAMPLE. — What  is  the  diameter  of  the  pitch  circle  of  a  gear-wheel 
which  has  75  teeth,  and  whose  pitch  is  1.675  inches  ? 

P  T 

SOLUTION.— Formula,         D  =  Tr^rrr^- 
6. 141  b 

Substituting,  we  have  D  =     '3'?4vL —  =  40  in.     Ans. 

To  find  the  number  of  teeth  in  a  gear-wheel  when  the 
diameter  and  pitch  are  given: 

Rule  87. —  The    number  of  teeth  equals   the  product  of 
3.1416  and  the  diameter  divided  by  the  pitch. 

rm,     *    •  ~  3.1416/? 

That  is,  /  = -p . 

EXAMPLE. — The  diameter  of  a  gear-wheel  is  40  inches,  and  the  pitch 
of  the  teeth  is  1.675  inches;  how  many  teeth  are  there  in  the  wheel  ? 


MECHANICS. 


179 


3. 1416  Z? 


l.Ot  u 


*  4°  =  75  teeth. 


Ans. 


SOLUTION.— Formula,          7 
Substituting,  we  have    T  = 

To  find  the  pitch  of  a  gear-wheel  when  the  diameter  and 
the  number  of  teeth  are  given. 

Rule  88. —  The  pitch  of  the  teeth  equals  the  product  of 
8.1416  and  the  diameter  divided  by  the  number  of  teeth. 

That  is,  P=  '  '       . 


EXAMPLE. — The  diameter  of  a  gear-wheel  is  40  inches,  and  it  has  75 
teeth;  what  is  the  pitch  of  the  teeth  ? 

SOLUTION.— Formula,         P  =  8'1416Z>. 


Substituting,  we  have  P  =  3-141r6x4°  _  1.675  in.  pitch. 

1 5 


Ans. 


509.  The  forms  of  teeth  used  in  ordinary  practice  are 
the  epicycloidal  and  involute. 

Fig.  88  shows  the  epicycloidal  form,  which  is  composed  of 
two  different  curves,  the  curve  from  the  pitch  circle  to  the 
top  of  the  tooth  being  an  epicycloid,  and  that  from  the  pitch 
circle  to  the  bottom  of  the  tooth  being  a  hypocycloid. 

In  gear-wheels  where  this  form  of  tooth  is  employed, 
their  pitch  circles  must  run  tangent  to  one  another. 

510.  In  Fig.  89  is  shown  the  involute  form  of  teeth,  or 


(  FIG.  89. 

teeth  having  but   one   curve.     The   outlines  of  the   teeth 
shown  in  the  rack  are  formed  of  straight  lines. 


180  MECHANICS. 

Involute  teeth  have  two  great  advantages  over  epicycloidal 
teeth:  1.  They  are  stronger  for  the  same  pitch,  as  they 
are  thicker  at  the  root.  2.  They  may  be  spread  apart  so 
that  their  pitch  circles  do  not  run  tangent  to  one  another 
without  practically  affecting  the  perfect  action  of  the 
teeth. 

511.  To  calculate  the  number  of  teeth  or  speed  of  one 
of  two  gear-wheels  which  are  to  gear  together: 

Let  N  =  number  of  revolutions  per  minute  of  the  driver; 
n  =  number  of  revolutions  per  minute  of  the  driven; 
T  =  number  of  teeth  in  the  driver  ; 
/  —  number  of  teeth  in  the  driven. 

Rule  89.  —  The  number  of  teeth  in  the  driver  equals  the 
product  of  the  number  of  teeth  and  number  of  revolutions  of 
the  driven  divided  by  the  number  of  revolutions  of  the  driver. 

That  is,  T=W- 

EXAMPLE.  —  The  driven  has  27  teeth,  and  will  make  66  revolutions 
per  minute  ;  if  the  driver  makes  99  revolutions  per  minute,  how  many 
teeth  are  there  in  the  driver  ? 

SOLUTION.  —  Formula,  T=-^-. 

Substituting,  we  have  T=  —     ^  =  18  teeth.     Ans. 


The  number  of  revolutions  per  minute  of  the  driver  and 
driven  and  the  number  of  teeth  in  the  driver  being  given, 
to  find  the  number  of  teeth  in  the  driven  : 

Rule  9O.  —  The  number  of  teeth  in  the  driven  equals  the 
product  of  the  number  of  teeth  and  revolutions  per  minute 
of  the  driver  divided  by  the  number  of  revolutions  per 
minute  of  the  driven. 

That  is,  /=™ 

n 

EXAMPLE.—  The  driver  has  18  teeth,  and  makes  99  revolutions  per 
minute,  and  the  driven  must  make  66  revolutions  per  minute;  how 
many  teeth  must  there  be  in  the  driven  ? 


SOLUTION.  —  Formula,  /  = 

ft 


MECHANICS.  181 


Substituting,  we  have  /  =  =  27  teeth.     Ans. 

The  number  of  teeth  in  the  driver  and  driven  and  the 
number  of  revolutions  per  minute  of  the  driver  being  given, 
to  find  the  number  of  revolutions  per  minute  of  the  driven: 

Rule  91.  —  The  number  of  revolutions  per  minute  of  the 
driven  equals  the  product  of  the  number  of  teeth  and  number  of 
revolutions  of  the  driver  divided  by  the  number  of  teeth  of  the 
driven. 

That  is,  n  =  ™. 

EXAMPLE.  —  There  are  18  teeth  in  the  driver,  and  it  makes  99  revo- 
lutions per  minute;  how  many  revolutions  per  minute  will  the  driven 
make  if  it  has  27  teeth  ? 

T  N 
SOLUTION.  —  Formula,  n  —  -y-. 

Substituting,  we  have  n  =  18^"  =  66  R.  P.  M.     Ans. 

The  number  of  teeth  in  the  driver  and  driven  and  the 
number  of  revolutions  per  minute  of  the  driven  being  given, 
to  find  the  number  of  revolutions  per  minute  of  the  driver: 

Rule  92.  —  The  number  of  revolutions  of  the  driver  equals 
the  product  of  the  number  of  teeth  and  revolutions  of  the 
driven  divided  by  the  number  of  teeth  of  the  driver. 

That  is,  N=*-£. 

EXAMPLE.—  If  there  are  27  teeth  in  the  driven,  and  if  it  makes  66 
revolutions  per  minute,  how  many  revolutions  per  minute  will  the 
driver  make  if  it  has  18  teeth  ? 

SOLUTION.  —  Formula,         N=  -^. 

Substituting,  we  have  N=  —        -  =  99  R.  P.  M.    Ans. 


EXAMPLE.—  In  Fig.  90,  the  crank-shaft  makes  60  revolutions  per 
minute  ;  the  governor  pulley  is  4"  in  diameter,  the  bevel  gear  on  the 
governor  pulley  shaft  has  19  teeth  ;  the  bevel  gear  which  meshes  with 


182 


MECHANICS. 


it  and  drives  the  governor  has  30  teeth.  The  governor  is  to  make  95 
revolutions  per  minute ;  what  should  be  the  size  of  the  pulley  on  the 
crank-shaft  ? 


FIG.  90. 


SOLUTION. — First  determine  the  number  of  revolutions  of  the  4* 
pulley  in  order  that  the  governor  shall  turn  95  times  per  minute. 

/  n       ^0  "v  Q^J 
Applying  rule   92,   N=-^= — ^ —  =  150  revolutions  of  gear  on 

pulley  shaft  =  revolutions  of  governor  pulley.  Now,  applying  rule 
8 1 ,  the  diameter  of  the  pulley  on  the  crank-shaft  =  -jr=-  =  — ^ —  = 
10'.  Ans. 

EXAMPLE. — In  Fig.  90,  the  fly-wheel  is  8  feet  in  diameter  and  drives 
a  5-foot  pulley  on  the  main  shaft.  A  14"  pulley  on  the  main  shaft 
drives  a  16"  pulley  on  the  countershaft.  A  12"  pulley  on  the  counter- 
shaft drives  a  12"  pulley  on  a  shaft  on  which  is  a  pinion  that  meshes 
into  a  large  gear  attached  to  the  face  plate  of  a  large  lathe,  and  which 
has  108  teeth.  How  many  teeth  must  the  pinion  have  in  order  that 
the  face  plate  may  make  9£  revolutions  per  minute  ?  " 

SOLUTION.— Applying  rule  83,  to  find  the  revolutions  per  minute 

of  the  main  shaft,  — = — -  =  96  R.  P.  M.     Applying  the  same  rule  again 
o 

to  find  the  revolutions  of  the  countershaft,   U  *  96  =  84  R.  P.  M. 

ID 

Applying  it  once  more  to  find  revolutions  of  the  pulley  which  turns 
the  small  gear,  ^y|—  =  84  R.  P.  M.  Applying  rule  89,  108^  9i  =  12 
teeth  in  pinion  or  driver.  Ans. 


MECHANICS.  183 

EXAMPLES   FOR   PRACTICE. 

1.  The  driving  pulley  makes  110  R.  P.  M.  and  is  21"  in  diameter; 
what  should  be  the  size  of  the  driven  in  order  to  make  385  R.  P.  M.  ? 

Ans.  6'. 

2.  The  main  shaft  of  a  certain  shop  makes  120  R.  P.  M.     It  is 
desired  to  have  the  countershaft  make  150  R.  P.  M.     There  are  on 
hand  pulleys  of  16",  24",  28",  35",  and  38"  in  diameter.     Can  two  of  these 
be  used,  or  must  a  new  pulley  be  ordered  ? 

Ans.  Use  the  28"  and  35"  pulleys. 

3.  The  pinion  (driver)  makes  174  R.  P.  M.,  and  follower  makes  24 
R.  P.  M. ;  how  many  teeth  must  the  pinion  have  if  the  follower  has 
87  teeth  ?  Ans.  12  teeth. 

4.  If  an  engine  fly-wheel  is  66"  in  diameter,  and  makes  160  R.  P.  M., 
what  must  be  the  diameter  of  the  pulley  on  the  main  shaft  to  make 
128  R.  P.  M.?  Ans.  82*'. 

5.  What  is  the  pitch  diameter  of  a  gear  whose  pitch  is  l£",  and  has 
28  teeth?  Ans.  11.14". 

6.  How  many  teeth  are  there  in  a  gear  whose  pitch  is  .7854",  and 
which  is  23"  in  diameter  ?  Ans.  92  teeth. 

7.  What  is  the  pitch  of  a  gear  whose  diameter  is  20.372*,  and  which 
has  128  teeth  ?  Ans.  i". 

8.  In  a  train  of  gears  the  drivers  have  16,  30,  24,  and  18  teeth, 
respectively ;  the  followers  have  12,  24,  36,  and  40  teeth,  respectively. 
If  the  first  driver  makes  80  R.  P.  M.,  how  many  R.  P.  M.  wilt  the  last 
follower  make  ?  Ans.  40  R.  P.  M. 


FIXED    AND    MOVABLE    PULLEYS. 

512.  Pulleys  are  also  used  for  hoisting  or  raising  loads,  in 
which  case  the  frame  which  supports  the  axle 

of  the  pulley  is  called  the  block. 

513.  A  fixed  pulley  is  one  whose  block 
is  not  movable,  as  in  Fig.  91.      In  this  case  if 
the  weight  W  be  lifted  by  pulling  down  /-*,  the 
other  end  of  the  cord  W  will  evidently  move 
the  same   distance   upwards   that  P    moves 
downwards;  hence,  P  must  equal  W. 

514.  A  movable  pulley  is   one   whose 

block  is  movable,  as  in  Fig.  92.  One  end  of  the         FIO.  91. 
cord  is  fastened  to  the  beam,  and  the  weight  is  suspended 


184 


MECHANICS. 


from  the  pulley,  the  other  end  of  the  cord  being  drawn 
up  by  the  application  of  a  force  P.  A  little  con- 
sideration will  show  that  if  P  moves  through  a  certain 
distance,  say  1  foot,  W  will  move  through  half  that  dis- 
tance, or  6  inches;  hence,  a  pull  of  one  pound  at  P  will 
lift  2  pounds  at  W. 

The   same   would   also   be   true   if  the   free  end  of   the 

cord  were  passed  over  a 
fixed  pulley,  as  in  Fig.  93, 
in  which  case  the  fixed 
pulley  merely  changes  the 
direction  in  which  P  acts, 
so  that  a  weight  of  1 
pound  hung  on  the  free 
end  of  the  cord  will  bal- 
ance 2  pounds  hung  from 
the  movable  pulley. 

FIG.  92.  FIG.  93.  515.     A       combina- 

tion of  pulleys,  as  shown  in  Fig.  94,  is  sometimes  used.  In 
this  case  there  are  three  movable  and  three  fixed  pulleys,  and 
the  amount  of  movement  of  W,  owing  to  a 
certain  movement  of  Pt  is  readily  found. 

It  will  be  noticed  that  there  are  six 
parts  of  the  rope,  not  counting  the  free 
end;  hence,  if  the  movable  block  be  lifted 
1  foot,  P  remaining  in  the  same  position, 
there  will  be  1  foot  of  slack  in  each  of  the 
six  parts  of  the  rope,  or  six  feet  in  all. 
Therefore,  P  would  have  to  move  6  feet 
in  order  to  take  up  this  slack,  or  P  moves 
6  times  as  far  as  W.  Hence,  1  pound  at  P 
will  support  6  pounds  at  W,  since  the 
power  multiplied  by  the  distance  through 
•which  it  moves  equals  the  weight  multiplied 
by  the  distance  through  which  it  moves. 
It  will  also  be  noticed  that  there  are 
three  movable  pulleys,  and  that  3x2  =  6.  FIG 


MECHANICS. 


185 


516.  Law  of  Combination  of  Pulleys  : 
Rule  93. — In  any  combination  of  pulleys  where  one  con- 
tinuous rope  is  used,  a  load  on  the  free  end  will  balance  a 
weiglit  on  the  movable  block  as  many  times  as  great  as  itself 
as  there  are  parts  of  the  rope  supporting  the  load,  not  counting 
the  free  end. 

The  above  law  is  good,  whether  the  pulleys  are  side  by 
side,  as  in  the  ordinary  block  and  tackle,  or  whether  they 
are  arranged  as  in  the  figure. 

EXAMPLE. — In  a  block  and  tackle  having  five  movable  pulleys,  how 
great  a  force  must  be  applied  to  the  free  end  of  the  rope  to  raise  1,250 
pounds  ? 

SOLUTION. — Since  there  are  five  movable  pulleys,  there  must  be 
10  parts  of  the  rope  to  support  them.  Hence,  according  to  the  above 
law,  a  force  applied  to  the  free  end  will  support  a  load  10  times  as  great 

1  950 
as  itself,  or  the  force  =  --  =  125  Ib.     Ans. 


517.     An 

making  an   angle    with 
a  horizontal  line. 

Three  cases  may  arise 
in  practice  with  the  in- 
clined plane: 

1.  Where  the    power 
acts     parallel     to     the 
pkme,  as  in  Fig.  95. 

2.  Where    the 


THE    INCLINED    PLANE, 
inclined    plane    is  a  slope,  or  a  flat  surface, 


W 


Zlbs. 


FIG.  95. 

power   acts  parallel  to   the  base,  as  in 
Fig.  96. 

3.  Where  the  power 
acts  at  an  angle  to  the 
plane  or  to  the  base,  as 
in  Fig.  97. 

518.     In     Fig.  95 
the    relation    existing 
between  the  power  and 
FIO.  96.  the   weight    is    easily 


lib. 


186 


MECHANICS. 


found.      The  weight  ascends  a  distance  equal  to  c  b,  or   the 

height  of  the  inclined 
plane,  while  the  power 
descends  through  a 
distance  equal  to  a  b, 
or  the  length  of  the 
inclined  plane.  There- 
fore, 

Rule  94.—  TAe 
power  multiplied  by 
the  length  of  the  in- 

FIG.  9r.  dined  plane  equals  t/ie 

weight  multiplied  by  the  height  of  the  inclined  plane;   or, 
letting 

length  of  plane  =  /; 
base  of  plane  =  b\ 
height  of  plane  =  //; 
we  have 


p= 


,  ...     PI 
and  W=  -=-. 
n 


EXAMPLE.— The  length   of    an    inclined   plane   is  40  feet,  and  its 
height  is  5  feet;  what  force  P  will  sustain  a  weight  IV oi  100  lb.? 
W  h       100x5 
—  '-    -~^0~~ 


SOLUTION.—/*  = 


lb.     Ans. 


EXAMPLE.—  The  length  of  an  inclined  plane  is  8  feet,  and  its  height 
is  15  inches;  what  weight  will  120  lb.  support  ? 


SOLUTION.  —  lV=-  = 


=  768  lb.     Ans. 


In  Fig.  96  the  power  is  supposed  to  act  parallel  to  the 
base  for  any  position  of  W\  therefore,  while  W  is  moving 
from  the  level  a  c  to  #,  or  through  the  height  c  b  of  the 
inclined  plane,  P  will  move  through  a  distance  equal  to  the 
length  of  the  base  a  c.  Hence,  when  the  power  acts  parallel 
to  the  base,  we  have 

Rule  95.  —  The  power  multiplied  by  the  base  of  the  inclined 
plane  equals  the  weight  multiplied  by  the  height  of  the  plane, 

.  Wh  Pb 

whence,  P=  —  r—  and  W=-j-. 


MECHANICS.  187 

EXAMPLE.—  With  a  base  30  feet  long,  and  a  height  of  6  feet,  what 
power  will  sustain  a  weight  of  75  lb.? 


SOLUTION.—  P  =  =  15  lb.    Ans. 

OU 

EXAMPLE.  —  A  force  of  12  lb.  sustains  a  weight  on  a  plane  whose 
base  is  6  ft.  long,  and  height  18  inches.     Find  the  weight. 


SOLUTION.—  W  =     /"    = 48  lb-    Ans- 

For  Fig.  97  no  rule  can  be  given.  The  ratio  of  the  power 
to  the  weight  must  be  determined  by  trigonometry  for  every 
position  of  W;  for,  as  W  shifts  its  position,  the  angle  that 
its  cord  makes  with  the  face  of  the  plane  varies,  and  the 
magnitude  of  the  force  P  depends  on  this  angle;  the  smaller 
the  angle  the  less  is  P  required  to  be  to  support  a  given 
weight  on  a  given  plane. 

519.  The  -wedge  is  a  movable  inclined  plane,  and  is 
used  for  moving   a    great 

weight  a  short  distance. 
A  common  method  of 
moving  a  heavy  body  is 
shown  in  Fig.  98. 

Simultaneous  blows  of 
equal  force  are  struck  on 
the  heads  of  the  wedges, 
thus  raising  the  weight 

W.     The  laws  for  wedges  are  the  same  as  for  Case  2  of  the 
inclined  plane. 

THE    SCREW. 

520.  A  screw  is  a  cylinder  with  a  helical  projection 
winding  around  its  circumference.     This  helix  is  called  the 
thread    of    the    screw.      The    distance    that   a    point    on 
the   helix   is  drawn    back    or    advanced    in    the    direction 
of  the  length  of  the  screw  during  one  turn,  is  called  the 
pitch  of  the  screw. 


188 


MECHANICS. 


The  screw  in  Fig.  99  is  turned  in  a  nut  a,  by  means  of  a 

force  applied  at  the 
end  of  the  handle  P. 
For    one    complete 
revolution    of    the 
handle,     the     screw 
will      be     advanced 
lengthwise,    an 
amount  equal  to  the 
pitch.     If  the  nut  be 
fixed,  and  a  weight 
be  placed  upon  the 
end  of  the  screw,  as 
shown,    it    will    be 
raised  vertically  a  distance  equal   to 
the   pitch    by   one  revolution    of   the 
screw.     During   this   revolution,    the 
force  at  P  will  move  through  a  dis- 
tance   equal    to     the     circumference 
whose     radius    is   P   F.      Therefore, 
W  X  pitch    of     thread  =  P  X  circum- 
ference of  P. 

521.     Hence,  to  find  the  weight 
that  a  given  force  will  lift,  we  have 

Rule  96. — Multiply   the  force  by  its  distance  from   the 
axis,  or  center  of  screw,  and  by  6. 2832 ';  divide  this  product 
by  the  pitch,  and  the  quotient  will  be  the  weight  required; 
PX  6.2832  X  r 


or 


whence, 


P= 


6.2832  X  r 


EXAMPLE.— It  is  desired  to  raise  a  weight  by  means  of  a  scre.w 
having  5  threads  per  inch.  The  force  is  40  pounds,  and  is  applied  at  a 
distance  of  14  inches  from  the  center  of  the  screw  (FP,  in  Fig.  99);  how 
great  a  weight  can  be  raised  ? 

SOLUTION.— Applying  the  formula, 

...     40x6.2832x14 

W= j —     —  =  17593  lb.,  very  nearly.     Ans. 


MECHANICS.  189 

EXAMPLE.— The  weight  to  be  lifted  is  4,000  pounds;  the  pitch  of 
the  screw  is  ^  (that  is,  there  are  4  threads  per  inch).  The  force  is 
applied  at  a  distance  of  12  inches  from  the  center  of  screw;  how  great 
must  it  be  ? 

SOLUTION.— Applying  the  formula, 

P  -  g4^^  X  *3  =  rather  more  than  13J  Ib. ,  nearly.     Ans. 

522.  Single-threaded  screws  of  less  than  1  inch  pitch 
are  generally  classified  by  the  number  of  threads  they  have 
in  1  inch  of  their  length.     In  such  cases  one  inch  divided  by 
the  number  of  threads  equals  the  pitch;  thus,  the  pitch  of  a 
screw  that  has  8  threads  per  inch  is£;  one  of  32  threads 
per  inch  is  ^,  etc. 

523.  Velocity    Ratio.— The    ratio    of    the    distance 
through  which  the  power  moves  to  the  corresponding  dis- 
tance   through    which    the    weight    moves   is    called    the 
velocity  ratio. 

Thus,  if  the  power  move  through  12  inches  while  the 
weight  moves  through  1  inch,  the  velocity  ratio  is  12  to  1,  or 
12;  that  is,  P  moves  12  times  as  fast  as  W. 

If  the  velocity  ratio  be  known,  the  weight  which  any 
machine  will  raise  can  be  found  by  multiplying  the  power 
by  the  velocity  ratio.  If  the  velocity  ratio  is  8.7  to  1,  or 
8.7,  W=  8.7  X  P,  since  W  X  1  =  Px  8.7. 

NOTE.— In  all  of  the  preceding  cases,  including  the  last,  the  effect 
of  friction  has  been  neglected. 


FRICTION. 

524.  Friction   is   the   resistance   that   a   body  meets 
with  from  the  surface  on  which  it  moves. 

525.  The  ratio  between  the  resistance  to  the  motion 
of  a  body  due  to  friction  and  the  perpendicular  pressure 
between  the  surfaces  is  called  the  coefficient  of  friction. 

If  a  weight    W,  as  in  Fig.  100,  rests  upon  a  horizontal 
plane,  and  has  a  cord  fastened  to  it  passing  over  a  pulley 

a,  from  which  a  weight  P  is  suspended,  then,  if  P  is  just 

p 
sufficient   to  start    W^  the   ratio  of  P  to   W,  or  -,  is  the 


190  MECHANICS. 

coefficient  of  friction  between  PFand  the  surface  upon  which 
it  slides. 

The  weight  W\s>  the  perpendicular  pressure,  and  Pis  the 


100  Ibs. 
W 


10  Iba. 


force  necessary  to  overcome  the  resistance  to  the  motion  of 
W,  due  to  friction. 
If  W=.  100  pounds  and  P  —  10  pounds,  the  coefficient  of 

P        10 

friction  for  this  particular  case  would  be  -jp-=  — —  =  .1. 

W       100 

526.  Laws  of  Friction  : 

1.  Friction  is  directly  proportional  to  the  perpendicular 
pressure  between  the  two  surfaces  in  contact. 

2.  Friction  is  independent  of  the  extent  of  the  surfaces  in 
contact  when  the  total  perpendicular  pressure  remains  the 
same. 

3.  Friction  increases  -with  the  roughness  of  the  surfaces. 

4.  Friction    is   greater    between  surfaces  of  the    same 
material  than  between  those  of  different  materials. 

5.  Friction  is  greatest  at  the  beginning  of  motion. 

6.  Friction  is  greater  between  soft  bodies  than  between 
hard  ones. 

7.  Rolling  friction  is  less  than  sliding  friction. 

8.  Friction  is  diminished  by  polishing  or  lubricating  the 
surfaces. 

527.  Law  1  shows  why  the  friction  is  so  much  greater 
on  journals  after  they  begin  to  heat  than  before.     The  heat 
causes  the  journal  to  expand,  thus  increasing  the  pressure 
between   the   journal   and   its   bearing,  and,  consequently, 
increasing  the  friction, 


MECHANICS. 


191 


Law  2  states  that,  no  matter  how  small  the  surface  may 
be  which  presses  against  another,  if  the  perpendicular  pres- 
sure is  the  same  the  friction  will  be  the  same.  Therefore,  large 
surfaces  are  used  where  possible,  not  to  reduce  the  friction, 
but  to  reduce  the  wear  and  diminish  the  liability  of  heating. 

528.  For  instance,  if  the  perpendicular  pressure  be- 
tween a  journal  and  its  bearing  is  10,000  pounds,  and  the 
coefficient  of  friction  is  .2,  the  amount  of  friction  is 
10,000  X  .2  =  2,000  pounds. 

Suppose  that  one-half  the  area  of  the  surface  of  the 
journal  is  80  square  inches,  then,  the  amount  of  friction  for 
each  square  inch  of  bearing  is  2,000  -4-  80  =  25  pounds. 

If  half  the  area  of  the  surface  had  been  160  square  inches, 
the  friction  would  have  been  the  same,  that  is,  2,000  pounds; 
but  the  friction  per  square  inch  would  have  been  2,000 
H-  160  =  12£  pounds,  just  one-half  as  much  as  before,  and  the 
wear  and  liability  to  heat  would  be  one-half  as  great  also. 

TABLE  OF  COEFFICIENTS  OF  FRICTION. 

TABLE    10. 


Description  of  Surfaces 
in  Contact. 

Disposition 
of  Fibers. 

State  of  the 
Surfaces. 

Coefficient 
of 
Friction. 

Oak  on  oak                   . 

Parallel 

Drv 

48 

LJI  y 

Oak  on  oak  

Parallel 

Soaped 

.16 

Wrought  iron  on  oak  

Parallel 

Dry 

.62 

Wrought  iron  on  oak  

Parallel 

Soaped 

.21 

Cast  iron  on  oak  

Parallel 

Dry 

.49 

Cast  iron  on  oak  

Parallel 

Soaped 

.19 

Wrought  iron  on  cast  iron  .  . 

Slightly 
Unctuous 

.18 

Wrought  iron  on  bronze  

— 

Slightly 
Unctuous 

.18 

Cast  iron  on  cast  iron 

Slightly 

.15 

Unctuous 

192  MECHANICS. 

529.  The  power  which  is  required  to  raise  a  weight,  or 
overcome  an  equal  resistance  in  any  machine,  is  thus  always 
greater  than  this  -weight  or  resistance  divided  by  the  velocity 
ratio  of  the  machine. 

Thus,  if  there  were  no  friction,  a  machine  whose  velocity 
ratio  was  5  would,  by  an  application  of  100  pounds  of  force, 
raise  a  weight  of  500  pounds. 

Now,  suppose  that  the  friction  in  the  machine  is  equiva- 
lent to  10  pounds  of  -force;  then,  it  would  take  110  pounds 
of  force  to  raise  500  pounds. 

If,  in  the  above  illustration,  friction  were  neglected,  110 
pounds  X  5  =  550  pounds,  or  the  weight  that  110  pounds 
would  raise  ;  but,  owing  to.  the  frictional  resistance,  it  only 
raised  500  pounds.  Therefore,  we  have  for  the  ratio 

RAA 

between  the  two,  —  =—  =  .91.     That  is,  500  :  550::  .91  :  1. 
5oO 

530.  Efficiency.  —  This   ratio  between  the  weight  ac- 
tually raised  and  the  power  multiplied  by  the  velocity  ratio, 
is  called  the  efficiency  of  the  machine. 

For  example,  if  the  weight  actually  raised  by  a  machine, 
say  a  screw,  is  1,600  pounds,  and  the  power  multiplied  by 
the  velocity  ratio  is  2,400  pounds,  the  efficiency  of  this 
machine  is 


EXAMPLE.—  In  a  machine  having  a  combination  of  pulleys  and 
gears,  the  velocity  ratio  of  the  whole  is  9.75.  A  force  of  250  pounds 
just  lifts  a  weight  of  1,626  pounds.  What  is  the  efficiency  of  the 
machine  ? 

SOLUTION.—  Efficiency  =       *t62.?  -,  =  .6671,  or  66.71*.     Ans. 

*OU  X  «*.  /o 

Since  the  total  amount  of  friction  varies  with  the  load,  it 
follows  that  the  efficiency  will  also  vary  for  different  loads. 

531.  If  the  efficiency  of  a  machine  is  known,  the  force 
actually  required  to  raise  a  given  load  may  be  found  by 
dividing  the  load  by  the  product  of  the  velocity  ratio  of  the 
machine  and  the  efficiency.  Thus,  if  a  certain  machine  has 


MECHANICS.  193 

a  velocity  ratio  of  10. 6,  and  its  efficiency  is  G0#,  the  force 
which  must  actually  be  applied  to  raise  a  load  of  840  pounds 
is  840  -=-  10.0  X  .00  =  840  ~  G.3C  =  132.1  pounds,  nearly.  If 
there  had  been  no  losses  through  friction,  etc.,  the  force 
required  would  have  been  840  -H  10.  G  =  70.25  pounds,  nearly. 
If  the  efficiency  is  known,  the  weight  which  a  certain  force 
will  raise  may  be  found  by  multiplying  together  the  force, 
velocity  ratio,  and  the  efficiency.  Thus,  if  a  certain  machine 
has  a  velocity  ratio  of  6£,  and  an  efficiency  of  78$,  a  force  of 
140  pounds  will  raise  a  weight  of  140  X  6£  X  .78  =  709,8 
pounds. 

532.  When  finding  the  force  necessary  to  overcome  the 
friction,  the  perpendicular  pressure  on  the  surface  considered 
must  always  be  taken.  In  order  to  find  (approximately) 
the  maximum  force  that  is  required  to  overcome  the  friction 
between  cross-head  and  guides,  we  have 

Rule  97. — Multiply  the  total  piston  pressure  by  the 
length  of  crank  and  by  the  coefficient  of  friction  and  divide 
by  the  length  of  main  rod; 

or,  letting     P  —  total  piston  pressure; 

/  =  length  of  main  rod; 

r  =  length  of  crank; 

c  =  coefficient  of  friction  between  cross- 
head  and  guides; 

F=  force  required  to  overcome  this 
friction, 

zr 
we  have  r  = 

EXAMPLE.— An  engine  whose  piston  is  16*  in  diameter  carries  a 
steam  pressure  of  80  Ib.  per  sq.  in.  If  the  crank  is  12"  long,  and  the 
connecting-rod  is  66"  long,  what  is  the  perpendicular  pressure  on  the 
guides  ?  The  coefficient  of  friction  for  this  case  being  12£,  what  force 
will  be  required  to  overcome  the  friction  ? 

SOLUTION.  —  Pressure  on  piston  =  16»  X  .7854  X  80  =  16.085  Ib. 
1('"S(.'(.  '  =  2,924. 55  Ib.  =  perpendicular  pressure.  Ans.  2.924.55  X 
.12  =  350.95  Ib.  =  force  required  u>  overcome  the  friction.  Ans. 


194  MECHANICS. 

EXAMPLES  FOR   PRACTICE. 

1.  How  great  a  force  must  be  applied  to  the  free  end  of  the  rope  of 
a  block  and  tackle  which  has  four  movable  pulleys,  to  raise  a  weight 
of  746  Ib.  ?  Ans.  93*  Ib. 

2.  An  inclined  plane  is  30  ft.  long  and  7  ft.  high ;  what  force  is 
required  to  roll  a  barrel  of  flour  weighing  196  Ib.  up  the  plane,  friction 
being  neglected  ?  Ans.  45. 7£  Ib. 

3.  The  distance  from  the  axis  of  a  screw  to  the  point  on  the  handle 
where  the  force  is  applied  is  12".     The  screw  has  8  threads  per  inch. 
What  force  is  necessary  to  raise  a  weight  of  1,248  Ib.  ? 

Ans.  2.07  Ib.,  nearly. 

4  In  example  3,  what  should  be  the  length  of  the  handle  to  raise  a 
weight  of  5,400  Ib.  by  the  application  of  a  force  of  20  Ib.? 

Ans.  5.371",  nearly. 

5.  What  is  the  velocity  ratio  (a)  in  example  3  ?  (b)  In  example  4  ? 

Ans    \  ^  603'  nearlv' 
Ans'  )  (6)  270. 

6.  An  engine  piston  is  24"  in  diameter.     If  the  steam  pressure  is 
93  Ib.  per  sq.  in. ;  the  length  of  the  connecting-rod,  8  ft.  4" ;  the  length 
of  crank,  20",  and  coefficient  of  friction,  14$,  what  is  (a)  the  greatest 
perpendicular  pressure  on  the  guides  ?    (b)  The  force  required  to  over- 
come the  friction  ?  .         (  (a)  8,414.46  Ib. 

Ans"  \  (b)  1,178  Ib. 

CENTRIFUGAL    FORCE. 

533.  If  a  body  be  fastened  to  a  string  and  whirled,  so 
as  to  give  it  a  circular  motion,  there  will  be  a  pull  on  the 
string  which  will  be  greater  or  less,  according  as  the  velocity 
increases  or  decreases.  The  cause  of  this  pull  on  the  string 
will  now  be  explained. 

Suppose  that  the  body  is  revolved  horizontally,  so  that  the 

^ action  of  gravity  upon  it  will  always  be 

-W^~~    *"  the  same.     According  to  the  first  law  of 
\          motion,  a  body  put  in    motion    tends   to 
move  in  a  straight  line,  unless  acted  upon 
by  some  other  force,  causing  a  change  in 
the  direction.     When  the  body  moves  in 
FIG.  101.  a  circle,  the  force  that  causes  it  to  move 

in  a  circle  instead  of  a  straight  line  is  exactly  equal  to  the 
tension  of  the  string.     If  the  string  were  cut,  the  pulling 


MECHANICS.  195 

force  that  drew  it  away  from  the  straight  line  would  be 
removed,  and  the  body  would  then  "  fly  off  at  a  tangent; " 
that  is,  it  would  move  in  a  straight  line  tangent  to  the 
circle,  as  shown  in  Fig.  101. 

Since,  according  to  the  third  law  of  motion,  every  action 
has  an  equal  and  opposite  reaction,  we  call  that  force  which 
acts  as  an  equal  and  opposite  force  to  the  pull  of  the  string 
the  centrifugal  force,  and  it  acts  away  from  the  center 
of  motion. 

534.  The  other  force,  or  tension,  of  the  string  is  called 
the  centripetal  force,  and   it  acts  towards  the  center  of 
motion.      It  is  evident  that  these  two  forces,  acting  in  oppo- 
site directions,  tend  to    pull    the  string  apart,   and,  if  the 
velocity  be  increased  sufficiently,  the  string  will  break.     It 
is  also  evident  that  no  body  can  revolve  without  generating 
centrifugal  force. 

535.  The  value  of  the  centrifugal  force,  expressed  in 
pounds,  of  any  revolving  body  is  calculated  by  the  following 
rule: 

Rule  98. —  T/ie  centrifugal  force  equals  the  continued 
product  of  .00034,  the  weight  of  the  body  in  pounds,  the 
radius  in  feet  (taken  as  the  distance  between  the  center  of 
gravity  of  the  body  and  the  center  about  which  it  revolves}, 
and  the  square  of  the  number  of  revolutions  per  minute. 

Let  F  =  centrifugal  force  in  pounds ; 

W=  weight  of  revolving  body  in  pounds; 

R  —  radius  in  feet  of  circle  described  by  center  of 

gravity  of  revolving  body; 
N=  revolutions  per  minute  of  revolving  body;  then 

F=  .00034   W  R  N\ 

536.  In  calculating  the  centrifugal  force  of  fly-wheels, 
it  is  the  usual  practice  to  consider  the  rim  of  the  wheel  only, 
and  not  take  the  arms  and  hub  of  the  wheel  into  account. 
In  this  case  R  would  be  taken  as  the  distance  between  the 
center  of  the  rim  and  the  center  of  the  shaft. 


196  MECHANICS. 

EXAMPLE. — What  would  be  the  centrifugal  force  developed  by  a 
cast-iron  fly-wheel,  whose  outside  diameter  was  10  feet,  width  of  face 
20  inches,  and  thickness  of  rim  6  inches,  turning  at  the  rate  of  80 
revolutions  per  minute  ?  Take  the  weight  of  a  cubic  inch  of  cast 
iron  as  .261  Ib. 

SOLUTION. — First  calculate  the  weight  of  the  rim.  The  diameter 
of  the  rim  =  10  X  12  =  120  inches;  the  diameter  of  the  circle  midway 
between  the  inside  and  outside  diameters  of  the  rim  =  120  —  6  = 
114  inches.  The  number  of  cubic  inches  in  the  rim  =  114  X  3.1416  X  20  X 
6  =  42,977  cubic  inches.  42,977  X  .261  =  11,217  pounds  =  weight. 

114 
Radius  =  =f  H-  12  =  4J  feet.     R.  P.  M.  =  80. 

Hence,  by  rule  98,  centrifugal  force  =  .00034  X  11,217  X  4|  X  802  = 
115,939  pounds.  Ans. 


SPECIFIC    GRAVITY. 

537.  The  specific  gravity  of  a  body  is  the  ratio  be- 
tween its  weight  and  the  weight  of  a  like  volume  of  water. 

538.  Since  gases  are  so  much  lighter  than  water,  it  is 
usual  to  take  the  specific  gravity  of  a  gas  as  the  ratio  be- 
tween the  weight  of  a   certain  volume  of  the  gas  and  the 
weight  of  the  same  volume  of  air. 

EXAMPLE.  —  A  cubic  foot  of  cast  iron  weighs  450  pounds;  what  is  its 
specific  gravity,  a  cubic  foot  of  water  weighing  62.5  pounds  ? 

450 
SOLUTION.—    --  =  7.2.     Ans. 


539.  The  specific  gravities  of  different  bodies  are  given 
in  printed  tables  ;  hence,  if  it  is  desired  to  know  the  weight 
of  a  body  that  can  not  be  conveniently  weighed,  calculate 
its  cubical  contents.  Multiply  the  specific  gravity  of  the 
body  by  the  weight  of  a  like  volume  of  water,  remembering 
that  a  cubic  foot  of  water  weighs  62.5  pounds. 

EXAMPLE.—  How  much  will  3,214  cubic  inches  of  cast  iron  weigh  ? 
Take  its  specific  gravity  as  7.21. 

SOLUTION.—  Since  1  cubic  foot  of  water  weighs  62.5  pounds,  3,214 
cubic  inches  weigh 

|i|l|  X  62.5  =  116.25  pounds. 
116.25x7.21  =  838.16  pounds.     Ans. 


MECHANICS.  197 

EXAMPLE.—  What  is  the  weight  of  a  cubic  inch  of  cast  iron  ? 
SOLUTION.—  X  7.21  =  .2608  pound.     Ans. 


NOTE.  —  One  cubic  foot  of  pure  distilled  water  at  a  temperature  of 
39.2°  Fahrenheit  weighs  62.42  pounds,  but  the  value  usually  taken  in 
making  calculations  is  62£  pounds. 

EXAMPLE.  —  What  is  the  weight  in  pounds  of  7  cubic  feet  of  oxygen  ? 
SOLUTION.  —  One  cubic  foot  of  air  weighs  .08073  lb.,  and  the  specific 
gravity  of  oxygen  is  1.1056  compared  with  air;  hence, 

.08073  X  1.1056  X  7  =  .62479  pound,  nearly. 


EXAMPLES  FOR  PRACTICE. 

1.  The  balls  of  a  steam-engine  governor  each  weigh  5  pounds.     If 
they  revolve  in  a  circle  whose  diameter  is  14"  at  the  rate  of  80  revolu- 
tions per  minute,  what  is  the  centrifugal  force  of  each  ball  ? 

Ans.  6.3471b.,  nearly. 

2.  The  rim  of  a  cast-iron  engine  fly-wheel  has  an  outside  diameter 
of  15  feet.     If  the'rim  is  8*  thick  and  12"  wide,  and  the  fly-wheel  makes 
40  R.  P.  M.,  what  is  the  centrifugal  force  of  the  rim  ?      Ans.  52,785  lb. 

3.  If  a  cubic  foot  of  a  certain  alloy  weighs  678  pounds,  what  is  its 
specific  gravity  ?  Ans.   10.848. 

4.  What  is  the  weight  of  (a)  12.4  cubic  inches  of  lead  ?    (6)  of  steel  ? 
(c)  of  aluminum  ?  f  (a)  5.0964  lb. 

Ans. -j  (t>)  3.52161b. 
(  (c)  1.1161b. 

5.  The  specific  gravity  of  an  alloy  of  lead  and  zinc  is  8.26;  what  is 
the  weight  of  a  cubic  foot  ?  Ans.  516.25  lb. 

WORK; 

540.  Work  is  the  overcoming  of  resistance  continually 
occurring  along  tJie  patli  of  motion. 

Mere  motion   is  not  work,  but  if  a  body  in  motion  con- 
stantly overcomes  a  resistance,  it  does  work. 

541.  The  measure  of  work  is  one  pound  raised  ver- 
tically one  foot,  and  is  called  one  foot-pound.     All  work 
is  measured  by  this  standard.     A  horse  going  up  hill  does 
an  amount  of  work  equal  to  his  own  weight  plus  the  weight 
of  the  wagon  and  contents  plus  the  frictional  resistances 
reduced  to  an  equivalent  weight  multiplied  by  the  vertical 
height  of  the  hill.     Thus,  if  the  horse  weighs  1,200  pounds, 
the  wagon  and  contents  1,200  pounds,  and  the  frictional 


198  MECHANICS. 

resistances  equal  400  pounds,  then,  if  the  vertical  height  of 
the  hill  is  100  feet,  the  work  done  is  equal  to  (1,200  -f  1,200 
-f  400)  X  100  =  280,000  foot-pounds. 

Rule  99.  —  In  all  cases  the  force  (or  resistance]  multiplied 
by  tJie  distance  through  which  it  acts  equals  the  work.  If  a 
weight  be  raised,  the  weight  multiplied  by  the  vertical  heig/tt 
of  the  lift  equals  the  work. 

542.  The  total  amount  of  work  is  independent  of  time, 
whether  it  takes  one  minute  or  one  year  in  which  to  do  it; 
but  in  order  to  compare  the  work  done  by  different  machines 
with  a  common  standard,  time  must  be  considered.     If  one 
machine  does  a  certain  amount  of  work  in  10  minutes,  and 
another  machine  does  exactly  the  same  amount  of  work  in 
5  minutes,  the  second  machine  can  do  twice  as  much  work 
as  the  first  in  the  same  time. 

543.  The  common  standard  to  which  all  work  is  reduced 
is  the  horsepower. 

One  horsepower  is  33,000  foot-pounds  per  minute  ;  in 
'other  words,  it  is  33,000  pounds  raised  vertically  one  foot  in 
one  minute,  or  1  pound  raised  vertically  33,000  feet  in  one 
minute,  or  any  combination  that  will,  when  multiplied  together, 
give  33,000  foot-pounds  in  one  minute. 

544.  Thus,  the  work  done  in  raising  110  pounds  verti- 
cally 5  feet  in  one  second  is  a  horsepower  ;  for,  since  in  one 
minute  there  are   60  seconds,    110  X  5  X  60=  33,000  foot- 
pounds in  one  minute. 

EXAMPLE.  —  If  the  coefficient  of  friction  is  .3,  how  many  horsepower 
will  it  require  to  draw  a  load  of  10,000  pounds  on  a  level  surface,  a  dis- 
tance of  one  mile  in  one  hour  ? 

SOLUTION.  —  10,000  X  -3  =  3,000  pounds  =  the  force  necessary  to  over- 
come the  resistance  (resistance  of  the  air  is  neglected).  One  mile  = 

pr   OQA 

5,280  feet;  one  hour  =  60  minutes.  Therefore,  -^-  =  88  feet  per 
minute. 

Work  done  =  force  multiplied  by  the  space  =  3,000  X  88  =  264,000 
foot-pounds  per  minute. 

264  000 
Horsepower  =  —     —  =  8  horsepower.     Ans. 


The  abbreviation  for  horsepower  is  H.  P. 


MECHANICS.  199 

545.  Energy  is  a  term  used  to  express  the  ability  of  an 
agent  to  do  work.     Work  can  not  be  done  without  motion, 
and  the  work   that   a  moving  body  is  capable  of  doing  in 
being  brought  to  rest  is  called  the  kinetic  energy  of  the 
body. 

Kinetic  energy  means  the  actual  visible  energy  of  a  body 
in  motion.  The  work  which  a  moving  body  is  capable  of 
doing  in  being  brought  to  rest  is  exactly  the  same  as  the 
kinetic  energy  developed  by  it  in  falling  in  a  vacuum  through 
a  height  sufficient  to  give  it  the  same  velocity. 

Rule  1OO.  —  The  kinetic  energy  of  a  moving  body  in  foot- 
pounds equals  its  weight  in  pounds  multiplied  by  tlie  square  of 
its  velocity  in  feet  per  second,  and  divided  by  64-32. 

If  W  represents  the  weight  of  the  body  in  pounds,  and  v 
the  velocity  in  feet  per  second, 

Wv* 
kinetic  energy  =  -^-^. 

If  a  weight  is  raised  a  certain  height,  a  certain  amount  of 
work  is  done  equal  to  the  product  of  the  weight  and  the 
vertical  height.  If  a  weight  is  suspended  at  a  certain  height 
and  allowed  to  fall,  it  will  do  the  same  amount  of  work  in 
foot-pounds  that  was  required  to  raise  the  weight  to  the 
height  through  which  it  fell. 

EXAMPLE.—  If  a  body  weighing  25  pounds  falls  from  a  height  of  100 
feet,  how  much  work  can  it  do  ? 

SOLUTION.—  Work  =  W  '  h  —  25  x  100  =  2,500  foot-pounds.     Ans. 

546.  It  requires  the  same  amount  of  work  or  energy  to 
stop  a  body  in  motion  within  a  certain  time  as  it  does  to  give 
it  that  velocity  in  the  same  time. 

EXAMPLE.—  A  body  weighing  50  pounds  has  a  velocity  of  100  feet  per 
second;  what  is  its  kinetic  energy? 


SoLUTioN.-Kinetic  energy  =  -  =      &JT   =  7'773'63  foot' 

pounds.     Ans. 

EXAMPLE.—  In    the   last  example,  how  many  horsepower   will  be 
required  to  give  the  body  this  amount  of  kinetic  energy  in  3  seconds  ? 
SOLUTION.—  1  H.  P.  =  83,000  pounds  raised  1  foot  in  one  minute. 
If  7,773.63  foot-pounds  of  work  are  done  in  3  seconds,  in  one  second 


200  MECHANICS. 

there  would  be  done  — — ^ —  =  2,591.21  foot-pounds  of  work.     One 
o 

horsepower  =  33,000  ft.-lb.  per  min.  =  33,000  -*-  60  =  550  ft.-lb.  per  sec. 
The  number  of  horsepower  developed  will  be 
2.591.21 


550 


=  4. 71 13  H.  P.     Ans. 


547".  Potential  energy  is  latent  energy ;  it  is  the  en- 
ergy which  a  body  at  rest  is  capable  of  giving  out  under  certain 
conditions. 

If  a  stone  is  suspended  by  a  string  from  a  high  tower,  it 
has  potential  energy.  If  the  string  is  cut,  the  stone  will  fall  to 
the  ground,  and  during  its  fall  its  potential  energy  will  change 
into  kinetic  energy,  so  that  at  the  instant  it  strikes  the  ground 
its  potential  energy  is  wholly  changed  into  kinetic  energy. 

At  a  point  equal  to  one-half  the  height  of  the  fall,  the  po- 
tential and  kinetic  energies  are  equal.  At  the  end  of  the 
first  quarter  the  potential  energy  was  £ ,  and  the  kinetic  en- 
ergy \;  at  the  end  of  the  third  quarter  the  potential  energy 
was  \,  and  the  kinetic  energy  f . 

A  pound  of  coal  has  a  certain  amount  of  potential  energy. 
When  the  coal  is  burned,  the  potential  energy  is  liberated 
and  changed  into  kinetic  energy  in  the  form  of  heat.  The 
kinetic  energy  of  the  heat  changes  water  into  steam,  which 
thus  has  a  certain  amount  of  potential  energy.  The  steam 
acting  on  the  piston  of  an  engine  causes  it  to  move  through 
a  certain  space,  thus  overcoming  a  resistance,  changing  the 
potential  energy  of  the  steam  into  kinetic  energy,  and  thus 
doing  work. 

Potential  energy  tJien,  is  the  energy  stored  within  a 
body  which  may  be  liberated  and  produce  motion,  thus 
generating  kinetic  energy,  and  enabling  work  to  be  done. 

548.  The  principle  of  conservation  of  energy  teaches 
that  energy,  like  matter,  can  never  be  destroyed.  If  a  clock 
is  put  in  motion,  the  potential  energy  of  the  spring  is  changed 
into  kinetic  energy  of  motion,  which  turns  the  wheels,  thus 
producing  friction. 

The  friction  produces  heat,  which  dissipates  into  the  sur- 
rounding air,  but  still  the  energy  is  not  destroyed — it  merely 


MECHANICS.  201 

exists  in  another  form.  The  potential  energy  in  coal  was 
received  from  the  sun,  in  the  form  of  heat,  ages  ago,  and  has 
lain  dormant  for  millions  of  years. 


BELTS. 

549.  A  belt  is  a  flexible  connecting  band  which  drives 
a  pulley  by  its  frictional  resistance  to  slipping  at  the  surface 
of  the  pulley.      Belts  are  most  commonly  made  of  leather  or 
rubber,  and  united  in  long  lengths  by  cementing,  riveting,  or 
lacing. 

550.  Belts  are  made  single  and  double.     A  single  belt 

is  one  composed  of  a  single  thickness  of  leather;  a  double 
belt  is  one  composed  of  two  thicknesses  of  leather  cemented 
and  riveted  together  the  whole  length  of  the  belt. 

551.  To  find  the  length  of  a  belt: 

In  practice,  the  necessary  length  for  a  belt  to  pass  around 
pulleys  that  are  already  in  their  position  on  a  shaft  is 
usually  obtained  by  passing  a  tape  line  around  the  pulleys, 
the  stretch  of  the  tape  line  being  allowed  as  that  necessary 
for  the  belt.  The  lengths  of  open-running  belts  for  pulleys 
not  in  position  can  be  obtained  as  follows: 

Rule  1O1. —  The  length  of  a  belt  for  open-running  pulleys 
equals  3^  times  one-half  the  sum  of  the  diameters  of  the  Al- 
leys plus  2  times  the  distance  between  the  centers  of  the  shaft. 
Let  D  =  diameter  of  one  pulley; 
-£>,=  diameter  of  other  pulley; 
L  =  distance  between  the  centers  of  the  shafts; 
B  =  length  of  the  belt. 

Then,  £  =  3±(D  +  D*\  +  2L. 

EXAMPLE. — The  distance  between  the  centers  of  two  shafts  is  9  feet 
7  inches;  the  diameter  of  the  large  pulley  is  36  inches,  and  the  diameter 
of  the  small  one  is  14  inches;  what  is  the  necessary  length  of  the  belt  ? 

SOLUTION.— By  formula,  B  =  3*  (/?^/?>)  +  2  L- 
Substituting  the  values  given,  we  have,  since  9  ft.  7'  =  115*, 

+  2  x  115  =  311±  in. .  or  25  ft.  llj  in.     Ans. 


202  MECHANICS. 

552.  To  find  the  width  of  a  single  leather  belt  that  will 
transmit  any  given  horsepower  when  equal  pulleys  are  used  : 

Rule  1O2.  —  The  width  of  the  belt  in  inches  equals  800 
times  the  horsepower  to  be  transmitted  divided  by  the  speed  of 
the  belt  in  feet  per  minute. 

Let  W  =  width  of  belt  in  inches; 

H  =  horsepower  to  be  transmitted  ; 
S  =  speed  of  belt  in  feet  per  minute;  then, 


EXAMPLE.  —  What  width  of  single  leather  belt  is  required  to  transmit 
20  horsepower  when  equal  pulleys  are  used,  and  the  speed  is  1,600  feet 
per  minute  ? 

SOLUTION.—  Formula,  W—  —  -<-=  —  . 

Substituting,  we  have 

.      800  X  20 

i  eon    —  10  inches.    Ans. 

553.  To  find  the  number  of  horsepower  that  a  single 
leather  belt  will  transmit,  its  width  and  speed  being  given  : 

Rule  1O3.  —  The  number  of  horsepower  equals  the  product 
of  the  width  in  inches  and  the  speed  in  feet  per  minute  divided 
by  800. 

-=^- 

EXAMPLE.  —  If  a  10"  single  leather  belt  is  to  be  run  at  a  speed  of 
1,600  feet  per  minute,  what  horsepower  will  it  transmit  ? 

W  S 
SOLUTION.—  Formula,  H  =  -^^-. 

oUv 

Substituting,  we  have 

H  =  10  ^J)'600  =  20  horsepower.     Ans. 

554.  When  the  pulleys  are  of  different  diameters,  the 
arc  of  contact  must  be  considered.     To  find  the  number  of 
degrees  in   the   arc  of  contact,  multiply  the  length  of  belt 
in  contact  on  the  smaller  pulley  by  360,  and  divide  the  product 
by  the  circumference  of  the  pulley,  calculating  the  result  to 
the  nearest  whole  number.    The  quotient  is  the  arc  of  contact. 


MECHANICS.  203 

Having  found  the  arc  of  contact,  subtract  it  from  180° 
and  multiply  the  result  by  3.  Add  this  last  result  to  800 ; 
the  number  thus  obtained  should  be  used  instead  of  800,  in 
rules  102  and  103. 

EXAMPLE.— What  should  be  the  width  of  a  single  leather  belt  to 
transmit  25.24  horsepower  at  a  speed  of  1,500  feet  per  minute,  the 
diameter  of  the  smaller  pulley  being  24",  and  the  belt  having  30"  of 
its  length  in  contact  with  it  ? 

SOLUTION.— Arc  of  contact  =  24°^^6  =  143°.     (180  -  143)  x  3  = 
111.     800  +  111  =  911.     Using  rule  1O2,  and  911  instead  of  800, 
.     911  X  25.24 


1,500 


=  15.33",  say  15f.     Ans. 


555.  To  find  the  width  of  a  double  belt  that  will  trans- 
mit the  same  horsepower  as  a  given  single  belt,  let  W^  repre- 
sent the  width  of  the  double  belt ;  then, 

Rule  1O4. — Multiply  the  width  of  a  single  belt  that  will 
transmit  the  same  horsepower  by  f . 

Or,  }\\  =  %W. 

EXAMPLE.— If  a  single  leather  belt  is  15*  in  width,  and  transmits 
22  horsepower,  what  must  be  the  width  of  a  double  belt  to  transmit 
the  same  horsepower  ? 

SOLUTION. — Applying  the  rule,  15  X  f  =  10  in.  =  width  of  double 
belt.  Ans. 

556.  Lacing  Belts. — Many  good  methods  of  fasten- 
ing the  ends  of  belts  are  employed,  but  lacing  is  generally 
used,  as  it  is  flexible,  like  the  belt,  and  runs  noiselessly  over 
the  pulleys. 

When  punching  a  belt  for  lacing,  use  an  oval  punch,  the 
long  diameter  of  the  hole  to  be  parallel  with  the  side  of  the  belt. 

In  a  3-inch  belt  there  should  be  four  holes  in  each  end, 
two  in  each  row.  In  a  6-inch  belt,  seven  holes,  four  in  the 
row  nearest  the  end.  A  10-inch  belt  should  have  nine  holes, 
five  in  the  row  nearest  the  end.  The  edges  of  the  holes 
should  not  be  nearer  than  £  of  an  inch  from  the  sides,  and 
$•  of  an  inch  from  the  ends  of  the  belt.  The  second  row 
should  be  at  least  If  inches  from  the  end. 

Always  begin  to  lace  from  the  center  of  the  belt,  and  take 


204 


MECHANICS. 


care  to  get  the  ends  exactly  in  line.  The  lacing  should  not 
be  crossed  on  the  side  of  the  belt  that  runs  next  to  the 
pulley.  Always  run  the  hair  side  of  the  belt  next  to  the 
pulley. 

HORSEPOWER    OF    GEARS. 

557.  To  find  the  horsepower  which  can  be  safely  trans- 
mitted by  gears  whose  face,  or  breadth  of  tooth,  is  from 
2£  to  3  times  their  pitch  : 

Rule  1O5. —  The  horsepower  which  can  be  safely  trans- 
mitted equals  the  continued  product  of  the  square  of  tlic  pitch, 
the  velocity  in  feet  per  minute  and  .01. 

Let  p  =  the  pitch ; 

s  =  circumferential   speed   of   a  point   on  the   pitch 
circle  in  feet  per  minute;  then, 
H.  P.  =.01  sp\ 

EXAMPLE.— What  horsepower  can  be  safely  transmitted  by  a  gear 
whose  pitch  diameter  is  66.84  in.,  pitch  If  in.,  and  which  makes 
60  R.  P.  M.? 

SOLUTION. — The  velocity  which  is  to  be  used  when  applying  rule 
1O5  is  the  circumferential  speed  of  a  point  on  the  pitch  circle. 


FIG.  102. 

Hence,    66.84  X  3.1416  =  209.98  in.  =  circumference  of  pitch  circle  = 
20Q  QR  20Q  Qft 

=^p  ft.      £~^  X  60  =  1,049.9  =  velocity  in  ft.  per  min. 

Now,  applying  formula,  H.  P.  =  .01  X  1,049.9  X  1.75*  =  32.1532  horse- 
power.    Ans. 


MECHANICS.  205 

558.  When  measuring  bevel  gears,  the  diameter  of  the 
largest  pitch  circle  should  be  taken,  as  D,  Fig.  102. 

When  calculating  their  horsepower,  use  the  small,  or  inner, 
diameter,  as  d,  Fig.  102.  Either  diameter  rr.ay  be  used  when 
calculating  the  revolutions  per  minute  or  number  of  teeth, 
by  rules  86-92,  but  if  the  inner  or  outer  diameter  of  one  gear 
be  used,  the  corresponding  diameter  of  the  other  gear  which 
meshes  with  it  must  also  be  used. 


EXAMPLES  FOR  PRACTICE. 

1.  How  many  foot-pounds  of  work  are  required  to  overcome  for 
7  minutes  the  friction  of  the  cross-head  of  an  engine  which  has  a  stroke 
of  4  ft.,  and  makes  160  strokes  per  minute,  if  the  coefficient  of  friction 
is  8$,  and  the  average  perpendicular  pressure  is  12,460  Ib.  ? 

Ans.  4,465,664  ft.-lb. 

2.  In  the  above  example,  what  horsepower  is  required  ? 

Ans.   1 9.332  H.  P. 

3.  A  cannon  ball  weighing  500  Ib.  is  fired  with  a  velocity  of  1,600ft. 
per  sec. ;  what  is  its  kinetic  energy  ?  Ans.   19,900,497.5  ft.-lb. 

4.  An  open  belt  drives  two  pulleys  which  are,  respectively,  42"  and 
20"  in  diameter,  and  23  ft.  apart  between  their  centers;  what  should  be 
the  length  of  the  belt  ?  Ans.  652J".  or  54  ft.  4f '. 

5.  What  width  of  single  leather  belting,  which  has  2  ft.  9"  contact 
on  the  small  pulley,  is  required  to  transmit  10  horsepower  at  a  speed  of 
1,500  ft.   per  min.?     Give  width    to  nearest  half  inch.     Diameter  of 
small  pulley  26".  Ans.  6". 

6.  What  should  be  the  width  of  the  main  belt  of  a  steam  engine  to 
transmit  120  horsepower  ?     The  engine  runs  at  80  R.  P.  M.,  the  band- 
wheel  is  8  ft.-  in  diameter,  the  belt  is  double,  and  has  a  contact  of  6  ft. 
on  the  smaller  pulley,  which  is  5  ft.  in  diameter.     Take  the  speed  of 
the  belt  the  same  as  that  of  a  point  on  the  circumference  of  the  band- 
wheel.  Ans.  36r. 

7.  A  26"  double  belt  runs  at  a  speed  of  2,830  ft.  per  min.,  and  has  a 
contact  of  5  ft.  on  the  smaller  pulley ;   what  horsepower  is  it  trans- 
mitting ?     Diameter  of  small  pulley  is  48".  Ans.  121.15  H.  P. 

8.  What  horsepower  can  be  safely  transmitted  by  a  gear  whose 
pitch  is  2*',  pitch  diameter  44.66",  and  which  makes  80  K.  P.  M.? 

Ans.  42.24  H.  P. 


MECHANICS. 

(CONTINUED.) 

HYDROSTATICS. 

559.  Hydrostatics  treats  of  liquids  at  rest  under  the 
action  of  forces. 

560.  Liquids  are  very  nearly  incompressible.     A  pres- 
sure of  15  pounds  per  square  inch  compresses  water  less  than 
26006  of  its  volume. 

561.  Fig.    103    represents    two   cylindrical   vessels  of 
exactly  the  same  size.     The  ves- 
sel a  is  fitted  with  a  wooden  block 

of  the  same  size  as  the  cylinder, 
and  can  move  in  it  ;  the  vessel 
b  is  filled  with  water,  whose 
depth  is  the  same  as  the  length 
of  the  wooden  block  in  a.  Both 
vessels  are  fitted  with  air-tight 
pistons  P,  whose  areas  are  each 
10  sq.  in. 

Suppose,  for  convenience,  that 
the  weights  of  the  pistons,  block, 
and  water  be  neglected,  and  that 
a  force  of  100  pounds  be  applied 
to  both  pistons.  The  pressure 


FIG.  103. 


100 


per  square  inch  will  be  — —  =  10  pounds.     In  the  vessel  a 

this  pressure  will  be  transmitted  to  the  bottom  of  the  vessel, 
and  will  be  10  pounds  per  square  inch ;  it  is  easy  to  see  that 
there  will  be  no  pressure  on  the  sides.  In  the  vessel  b  an 
entirely  different  result  is  obtained.  The  pressure  on  the 
bottom  will  be  the  same  as  in  the  other  case,  that  is, 
10  pounds  per  square  inch,  but,  owing  to  the  fact  that  the 


208 


MECHANICS. 


molecules  of  the  water  are  perfectly  free  to  move,  this  pres- 
sure of  10  pounds  per  square  inch  is  transmitted  in  every 
direction  with  the  same  intensity  ;  that  is  to  say,  the  pres- 
sure at  any  point  c,  d,  e,  f,  g,  h,  etc. ,  due  to  the  force  of 
100  pounds,  is  exactly  the  same,  and  equals  10  pounds  per 
square  inch. 

562.  This  may  be  easily  proven  experimentally  by 
means  of  an  apparatus  like  that  shown  in  Fig.  104.  Let  the 
area  of  the  piston  a  be  20 
sq.  in. ;  of  b,  7  sq.  in. ;  of 
c,  1  sq.  in. ;  of  d,  6  sq.  in. ; 
of  e,  8  sq.  in.,  and  of  /, 
4  sq.  in. 

If  the  pressure  due  to 
the  weight  of  the  water 
be  neglected,  and  a  force 
of  5  pounds  be  applied  at 
c  (whose  area  is  1  sq.  in.), 
a  pressure  of  5  pounds  per 
square  inch  will  be  trans- 
mitted in  all  directions, 
and  in  order  that  there 
shall  be  no  movement,  a 
force  of  6  X  5  —  30  pounds 
must  be  applied  at  d,  40  pounds  at  e,  20  pounds  at  /, 
100  pounds  at  a,  and  35  pounds  at  b. 

If  a  force  of  99  pounds  were  applied  to  a,  instead  of 
100  pounds,  the  piston  a  would  rise,  and  the  other  pistons 
$,  c,  d,  e,  and  f,  would  move  inwards  ;  but,  if  the  force 
applied  to  a  were  100  pounds,  they  would  all  be  in  equilib- 
rium. If  101  pounds  were  applied  at  a,  the  pressure  per 

square   inch  would  be— —  =  5.05  pounds,  which  would  be 

transmitted  in  all  directions  ;  and,  since  the  pressure  due  to 
the  load  on  c  is  only  5  pounds  per  square  inch,  it  is  now 
evident  that  the  piston  a  will  move  downwards,  and  the 
pistons  b,  c,  d,  c,  and/" will  be  forced  outwards. 


FIG.  104. 


MECHANICS. 


563.  The  whole  of  the  preceding  may  be  summed  up 
as  follows  : 

The  pressure  per  unit  of  area  exerted  anywhere  upon  a  mass 
of  liquid  is  transmitted  undiminished  in  all  directions,  and 
acts  with  the  same  force  upon  all  surfaces,  in  a  direction  at 
right  angles  to  those  surfaces. 

This  law  was  first  discovered  by  Pascal,  and  is  the  most 
important  in  hydromechanics.  Its  meaning  should  be 
thoroughly  understood. 

EXAMPLE. — If  the  area  of  the  piston  e,  in  Fig.  104,  were  8.25  sq.  in., 
and  a  force  of  150  pounds  were  applied  to  it,  what  forces' would  have  to 
be  applied  to  the  other  pistons  to  keep  the  water  in  equilibrium, 
assuming  that  their  areas  were  the  same  as  given  before  ? 

1  ^0 

SOLUTION. —    ITSF—  18-182  pounds  per  square  inch,  nearly. 
8.  £<) 

20  X  18.182  =  363. 64   Ib.  =  force  to  balance  a. 
7  X  18.182  =  127.274  Ib.  =  force  to  balance  b. 
1  X  18.182  =    18.182  Ib.  =  force  to  balance  c.       Ans. 
6  X  18.182  =  109.092  Ib.  =  force  to  balance  d. 
4  X  18.182  =    72.728  Ib.  =  force  to  balance  /. 

564.  The  pressure  due  to  the  weight  of  a  liquid 

may  be  downwards,  upwards,  or  sideways. 

565.  Downward  Pressure. — In  Fig.    105  the  pres- 

sure on  the  bottom  of  the  vessel 
a  is,  of  course,  equal  to  the 
weight  of  the  water  it  contains. 
If  the  area  of  the  bottom  of  the 
vessel  £,  and  the  depth  of  the 
liquid  contained  in  it,  are 
the  same  as  in  the  vessel  a, 
the  pressure  on  the  bottom 
of  b  will  be  the  same  as  on  the 
bottom  of  a.  Suppose  the  bot- 
toms of  the  vessels  a  and  b  are 
6  inches  square,  and  that  the 
FIG.  105.  part  c  d  in  the  vessel  b  is 

2  inches  square,  and  that  they  are  filled  with  water.     Then, 

the  weight  of  one  cubic  inch   of  water  is  p^  pound  = 


210  MECHANICS. 

.03617  pound.  The  number  of  cubic  inches  in  a  =  6  X  6 
X  24=864  cubic  inches;  and  the  weight  of  the  water  is 
864  X  .03617  =  31.25  pounds.  Hence,  the  total  pressure  on 
the  bottom  of  the  vessel  a  is  31.25  pounds,  or  0.868  pound 
per  square  inch. 

The  pressure  in  b,  due  to  the  weight  contained  in  the 
part  £<:is6x6xlOx  .03617  =  13.02  pounds. 

The  weight  of  the  part  contained  in  c  d  is  2  X  2  X  14 
X  .03617  =  2.0255  pounds,  and  the  weight  per  square  inch 

,  .    2.0255 
of  area  in  c  d  is  — - —  =  .5064  pound. 

566.  According  to  Pascal's  law,  this  weight  (pressure) 
is  transmitted  equally  in  all  directions,  therefore,  an  extra 
weight  of  .5064  pound  is  imposed  on  every  square  inch  of  the 
bottom  of  bc\  the  area  of  this  is  6  X  6  =  36  square  inches, 
and    the    pressure   on   it    is,   therefore,  36  X  .5064  =  18.23 
pounds,  due  to  water  contained  in  c  d;  thus,  we  have  a  total 
pressure   on   bottom   of   vessel   b  of   13.02+18.23  =  31.25 
pounds,  the  same  as  in  vessel  a.     As  a  result  of  the  above 
law,  there  is  also  an  upward  pressure  of  .5064  pound  acting 
on  every  square  inch  of  the  top  of  the  enlarged  part  be. 

If  an  additional  pressure  of  10  pounds  per  square  inch 
were  applied  to  the  upper  surface  of  both  vessels,  the  total 
pressure  on  each  bottom  would  be  31.25+  (6  X  6  X  10)  = 
31.25  +  360  =  391.25  pounds. 

If  this  pressure  were  to  be  obtained  by  means  of  a  weight 
placed  on  each  piston  (as  shown  in  Figs.  103  and  104),  we 
should  have  to  put  a  weight  of  6  X  6  X  10  =  360  pounds  on 
the  piston  in  vessel  a,  and  one  of  2  X  2  X  10  =  40  pounds 
on  the  piston  in  vessel  b. 

567.  The     General     Law    for     the     Downward 
Pressure  upon  the  Bottom  of  any  Vessel : 

Rule  1O6. —  The  pressure  upon  the  bottom  of  a  vessel  con- 
taining a  fluid  is  independent  of  the  shape  of  the  vessel,  and 
is  equal  to  the  weight  of  a  column  of  the  fluid,  the  area  of 
whose  base  is  equal  to  that  of  the  bottom  of  the  vessel,  and 
whose  altitude  is  the  distance  between  the  bottom  and  the 
upper  surface  of  the  fluid  plus  the  pressure  per  unit  of  area 


MECHANICS. 


211 


upon  the  upper  surface  of  the  fluid  multiplied  by  the  area  of 
the  bottom  of  the  vessel. 

568.  Suppose  that  the  vessel  b,  in  Fig.  105,  were  in- 
verted, as  shown  in  Fig.  106,  the  pressure  upon  the  bottom 
will  still  be  0.8G8  pound  per  square  inch,  but 
it  will  require  a  weight  of  3,490  pounds  to 
be  placed  upon  a  piston  at  the  upper  surface 
to  make  the  pressure  on  the  bottom  391.25 
pounds,  instead  of  a  weight  of  40  pounds,  as 
in  the  other  case. 

EXAMPLE.— A  vessel  filled  with  salt  water,  having 
a  specific  gravity  of  1.03,  has  a  circular  bottom 
13  inches  in  diameter.  The  top  of  the  vessel  is  fitted 
with  a  piston  3  inches  in  diameter,  on  which  is  laid 
a  weight  of  75  pounds ;  what  is  the  total  pressure  on 
the  bottom,  if  the  depth  of  the  water  is  18  inches  ? 

SOLUTION. — Applying  the  rule,  the  weight  of  1  cubic  inch  of  the 
62.5  X  1-03 


water  is  - 


= .037254  Ib. 


1,728 

13  X  13  X  -7854  X  18  X  .037254  =  89.01  pounds  =  the  pressure  due  to 
the  weight  of  the  water. 

5 — ^- — p^-r  =  10.61  pounds  per  square  inch,  due  to  the  weight  on 
o  X  «  X  • '  854 

the  piston.        13  x  13  x  .7854  X  10.61  =  1,408.29  pounds. 
Total  pressure  =  1.408. 29  +  89.01  =  1,497.3  pounds.     Ans. 

569.  Upward  Pressure. —  In  Fig.  107 
is  represented  a  vessel  of  exactly  the  same 
size  as  that  shown  in  Fig.  106.  There  is 
no  upward  pressure  on  the  surface  c,  due 
to  the  weight  of  the  water  in  the  large  part 
c  d,  but  there  is  an  upward  pressure  on  c,  due 
to  the  weight  of  the  water  in  the  small  part 
b  c.  The  pressure  per  square  inch,  due  to  the 
weight  of  the  water  in  b  c,  was  found  to  be 
.5064  pound  (see  Art.  565),  the  area  of  the 
upper  surface  c  of  the  large  part  c  d  is 
(6  X  C)  -  (2  X  2)  =  36  -  4  =  32  sq.  in.,  and 
FIG.  107.  tne  total  upward  pressure,  due  to  the  weight 
of  the  water,  is  .5064  X  32  =  16.2  pounds. 


212  MECHANICS. 

If  an  additional  pressure  of  10  pounds  per  square  inch 
were  applied  to  a  piston  fitting  in  the  top  of  the  vessel,  the 
total  upward  pressure  on  the  surface  c  would  be  16.2  + 
(32  X  10)  =  336.2  pounds. 

57O.     General  Law  for  Upward  Pressure: 

Rule  1O7.  —  The  upward  pressure  on  any  submerged  hori- 
zontal surface  equals  the  weight  of  a  column  of  the  liquid 
whose  base  has  an  area  equal  to  the  area  of  the  submerged 
surface,  and  whose  altitude  is  the  distance  between  the  sub- 
merged surface  and  the  upper  surface  of  the  liquid  plus  the 
pressure  per  unit  of  area  on  the  upper  surface  of  the  fluid 
multiplied  by  the  area  of  the  submerged  surface. 

EXAMPLE. — A  horizontal  surface  6  inches  by  4  inches  is  submerged 
in  a  vessel  of  water  26  inches  below  the  upper  surface ;  if  the  pressure 
on  the  water  is  16  pounds  per  square  inch,  what  is  the  total  upward 
pressure  on  the  horizontal  surface  ? 

SOLUTION. — Applying  the  rule,  we  get  6  X  4x  26  x  .03617  =  22.57 
pounds  for  the  upward  pressure,  due  to  the  weight  of  the  water,  and 
6  X  4  X  16  =  384  pounds  for  the  upward  pressure,  due  to  the  outside 
pressure  of  16  pounds  per  square  inch. 

Therefore,  the  total  upward  pressure  =  384  +  22.57  =  406.57  pounds. 
Ans. 

Lateral  (Sideways)  Pressure. — Suppose  the 
top  of  the  vessel  shown  in  Fig.  108  is 
10  inches  square,  and  that  the  pro- 
jections at  a  and  b  are  1  inch  X  1  inch 
and  10  inches  long. 

The  pressure  per  square  inch  on  the 
bottom  of  the  vessel,  due  to  the  weight 
of  the  liquid,  is  1  X  1  X  18  X  the  weight 
of  a  cubic  inch  of  the  liquid. 

The  pressure  at  a  depth  equal  to  the 
distance    of   the    upper     surface    b   is 
1  X  1  X  17  X  the  weight  of  a  cubic  inch 
FIG.  108.  of  the  liquid. 

Since  both  of  these  pressures  are  transmitted  in  every 
direction,  they  are  also  transmitted  laterally  (sideways), 


MECHANICS.  213 

and  the  pressure  per  unit  of  area  on  the  projection  b  is  a  mean 
between  t/tc  two,  and  equals  1  x  1  X  174-  X  the  weight  of  a 
cubic  inch  of  the  liquid. 

To  find  the  lateral  pressure  on  the  projection  a,  imag- 
ine that  the  dotted  line  c  is  the  bottom  of  the  vessel,  then 
the  conditions  would  be  the  same  as  in  the  preceding  case, 
except  that  the  depth  is  not  so  great. 

The  lateral  pressure  per  sq.  in.  on  a  is  thus  seen  to  be 
1  X  1  X  1H  X  the  weight  of  a  cubic  inch  of  the  liquid. 

572.  General  Law  for  Lateral  Pressure  : 

Rule  1O8. —  The  pressure  upon  any  vertical  surface,  due  to 
the  weight  of  the  liquid,  is  equal  to  the  weight  of  a  column  of 
the  liquid  whose  base  has  the  same  area  as  the  vertical  surface, 
and  whose  altitude  is  the  depth  of  the  center  of  gravity  of  the 
vertical  surface  below  the  upper  surface  of  'the  liquid. 

Any  additional  pressure  is  to  be  added  as  in  the  previous  cases. 

EXAMPLE.— A  well  3  feet  in  diameter,  and  20  feet  deep,  is  filled  with 
water;  what  is  the  pressure  on  a  strip  of  the  wall  1  inch  wide,  the  top 
of  which  is  1  foot  from  the  bottom  of  the  well  ?  What  is  the  pressure 
on  the  bottom  ?  What  is  the  upward  pressure  per  square  inch,  2  feet 
6  inches  from  the  bottom  ? 

SOLUTION. — Applying  the  rule,  the  area  of  the  strip  is  equal  to 
its  length  (—  circumference  of  well)  multiplied  by  its  height.  The 
length  =  36  X  3.1416  =  113.1" ;  height  =  1" ;  hence,  area  of  strip  =  113.1 
X  1  =  113.1  sq.  in.  Depth  of  center  of  gravity  of  strip  =  (20  —  1)  ft.  -f- 
|  inch,  the  half  width  of  strip,  =  228^  inches.  Consequently,  the  total 
pressure  on  the  strip  =  113.1  X  228.5  X  .03617  =  934.75  Ib.  Ans. 

The  pressure  on  each  square  inch  of  the  strip  would  be  '  =  8.265 
pounds,  nearly. 

36  X  36  X  -7854  x  20  X  12  X  .03617  =  8,836  pounds  =  the  pressure  on 
the  bottom.  Ans. 

20  -  2.5  =  17.5.  1  x  17.5  X  12  X  .03617  =  7.596  pounds  =  the  upward 
pressure  per  square  inch,  2  feet  6  inches  from  the  bottom.  Ans. 

573.  The  effects  of  lateral  pressure  are  illustrated  in 
Fig.  109.     In  the  figure,  e  is  a  tall  vessel  having  a  stop-cock 
near  its  base,  and  arranged  to  float  upon  the  water,  as  shown. 
When  this  vessel  is  filled  with  water,  the  lateral  pressures  at 
any  two  points  of  the  surface  of  the  vessel  opposite  to  each 


214 


MECHANICS. 


other  are  equal.      Being  equal,  and  acting  in  opposite  direc- 
tions, they  destroy  each  Bother,  and  no  motion  can  result; 

but  if  the  stop-cock 
be  opened  there  will 
be  no  resistance  to 
the  pressure  acting 
on  that  part,  and  the 
water  will  flow  out; 
at  the  same  time,  the 
pressure  on  the  cor- 
responding part  of 
the  opposite  side  of 
the  vessel  remains, 
and  this,  being  no 
longer  balanced, 
causes  the  vessel  to 
FlG-  lc9-  move  backwards 

through  the    water  in    a  direction  opposite  to  that  of  the 
issuing  jet. 

574.  Since  the  pressure  on  the  bottom  of  a  vessel,  due 
to  the  weight  of  the  liquid,  is  dependent  only  upon  the  height 
of  the  liquid,  and  not  upon  the  shape  of  the  vessel,  it  follows 


FIG.  no. 

that  if  a  vessel  has  a  number  of  radiating  tubes  (see  Fig.  110) 
the  water  in  each  tube  will  be  on  the  same  level,  no  matter 


MECHANICS.  215 

what  may  be  the  shape  of  the  tubes.  For,  if  the  water  were 
higher  in  one  tube  than  in  the  others,  the  downward  pressure 
on  the  bottom,  due  to  the  height  of  the  water  in  this  tube, 
would  be  greater  than  that  due  to  the  height  of  the  water 
in  the  other  tubes.  Consequently,  the  upward  pressure 
would  also  be  greater;  the  equilibrium  would  be  destroyed, 
and  the  water  would  flow  from  this  tube  into  the  vessel,  and 
rise  in  the  other  tubes  until  it  was  at  the  same  level  in  all, 
when  it  would  be  in  equilibrium.  This  principle  is  expressed 
in  the  familiar  saying,  water  seeks  its  level. 

575.  The    above    principle    explains   why   city    water 
reservoirs  are  located  on  high  elevations,  and  why  water  on 
leaving  the  hose  nozzle  spouts  so  high. 

If  there  were  no  resistance  by  friction  and  air,  the  water 
would  spout  to  a  height  equal  to  the  level  of  the  water  in 
the  reservoirs.  If  a  long  pipe  were  attached  to  the  nozzle 
whose  length  was  equal  to  the  vertical  distance  between  the 
nozzle  and  the  level  of  the  water  in  the  reservoir,  the  water 
would  just  reach  the  end  of  the  pipe.  If  the  pipe  were  lowered 
slightly,  the  water  would  trickle  out. 

Fountains,  canal  locks,  and  artesian  wells  are  examples  of 
the  application  of  this  principle. 

EXAMPLE. — The  water  level  in  a  city  reservoir  is  150  feet  from  the 
level  of  the  street ;  what  is  the  pressure  of  the  water  per  square  inch 
on  the  hydrant  ? 

SOLUTION.— Apply  rule  1O6,  1  X  150  x  12  X  .03617  =  65.106  pounds 
per  square  inch. 

NOTE.— In  measuring  the  height  of  the  water  to  find  the  pressure 
which  it  produces,  the  vertical  height,  or  distance,  between  the  level 
of  the  water  and  the  point  considered  is  always  taken.  This  vertical 
height  is  called  the  head. 

The  weight  of  a  column  of  water  1  in.  square  and  1  ft.  high  is 
62.5  H-  144  =  .434  lb.,  very  nearly.  Hence,  if  the  depth  (head)  be  given, 
the  pressure  per  square  inch  may  be  found  by  multiplying  the  depth  in 
feet  by  .434.  The  constant  .434  is  the  one  ordinarily  employed  in 
practical  calculations. 

576.  In    Fig.  Ill,   let    the    area  of    the    piston    a  be 
1  square  inch,  of  b,  40  square  inches.     According  to  Pascal's 
law,  1  pound  placed  upon  a  will  balance  40  pounds  placed 
upon  b. 


216  MECHANICS. 

Suppose  that  a  moves  downwards  10  inches,  then,  10  cubic 
inches  of  water  will  be  forced  into  the  tube  b.  This  will  be 
distributed  over  the  entire  area  of  the 
tube  $,  in  the  form  of  a  cylinder  whose 
cubical  contents  must  be  10  cubic  inches, 
whose  base  has  an  area  of  40  square 

10 
inches,  and -whose  altitude  must  be  —  = 

—  of  an  inch ;  that  is,  a  movement  of 
4 

10  inches  of  the  piston  a  will  cause  a 
movement  of  \  of  an  inch  in  the  pis- 
ton b. 

FIG-  m-  Here  is  the  old  principle  of  machines: 

The  power  multiplied  by  the  distance  through  which  it 
moves  equals  the  weight  multiplied  by  the  distance  through 
•whicli  it  moves. 

Since,  if  1  pound  on  the  piston  a  represents  the  power  P, 
the  equivalent  weight  W  on  b  may  be  obtained  from  the 
equation,  W  X  i  =  P  X  10,  whence  W  —  40  P=  40  Ib. 

577.  Another  familiar  fact  is  also  recognized,  for  the 
velocity  ratio  of  Pto  W  is  10  :  £,  or  40;  and,  since  in  any 
machine  the  weight  equals  the  power  multiplied  by  the 
velocity  ratio,  W  —  Px  40,  and  when  P=l,  W  =  40. 

This  principle  is  made  use  of  in  the  hydraulic  press  rep- 
resented in  Fig.  112.  As  the  man  depresses  the  lever  O, 
he  forces  down  the  piston  a  upon  the  water  in  the  cylinder 
A.  The  water  is  forced  through  the  bent  tube  d  into  the 
cylinder  in  which  the  large  ram,  or  plunger,  C  works,  and 
causes  it  to  rise,  thus  lifting  the  platform  K,  and  compress- 
ing the  bales.  If  the  area  of  a  be  1  sq.  in.,  and  that  of  c  be 
100  sq.  in.,  the  velocity  ratio  will  be  100. 

If  the  length  of  the  lever  between  the  hand  and  the  fulcrum 
is  10  times  the  length  between  the  fulcrum  and  the  piston  a, 
the  velocity  ratio  of  the  lever  will  be  10,  and  the  total 
velocity  ratio  of  the  hand  to  the  piston  C  will  be  10  X  100 
=  1,000. 


MECHANICS.  217 

Hence,  a  force  of  100  pounds  applied  by  the  hand  will  raise 
100  X  1,000  =  100,000  pounds.      But,   if  the  average  move- 


112. 

ment  of  the  hand  per  stroke    is  10  inches,   it  will   require 

1   000 

'         =  100  strokes  to  raise  the  platform  1  inch,  and  it  is 
again  seen  that  what  is  gained  in  power  is  lost  in  speed. 

578.  Applications  of  this  principle  are  seen  in  the 
hydraulic  machines  used  for  forcing  locomotive  drivers  on 
their  axles,  etc.,  and  for  testing  the  strength  of  boiler  shells. 

EXAMPLE. — A  suspended  vertical  cylinder  is  tested  for  the  tightness 
of  its  heads  by  filling  it  with  water.  A  pipe  whose  inside  diameter  is 
i  of  an  inch,  and  whose  length  is  20  feet,  is  screwed  into  a  hole  in  the 
upper  head,  and  then  filled  with  water.  What  is  the  pressure  per 
square  inch  on  each  head,  if  the  cylinder  is  40  inches  in  diameter -°nd 
60  inches  long  ? 

SOLUTION.— Area  of  heads  =  40*  X  .7854  =  1,256.64  sq.  in. 

Pressure  per  square  inch  on  the  bottom  head,  due  to  the  weight  of 
the  water  in  the  cylinder,  =  1  X  60  X  .03617  =  3.17  pounds- 


218 


MECHANICS. 


(i)*  X  .7854  -  .04909  sq.  in.,  the  area  of  the  pipe. 

.04909  X  20  X  12  X  .03617  =  .426  pound  =  the  weight  of  water  in  the 
pipe  =  the  pressure  on  a  surface  area  of  .04909  sq.  in. 

The  pressure  per  square  inch,  due  to  the  water  in  the  pipe,  is  „  .„  „  x 

.426  =  8.68  pounds  per  square  inch  upon  the  upper  head.     Ans. 

The  total  pressure  per  square  inch  on  the  lower  head  is  8.68  +  2.17  = 
10.85  pounds.  Ans. 

EXAMPLE. — In  the  last  example,  if  the  pipe  be  fitted  with  a  piston 
weighing  £  of  a  pound,  and  a  5-pound  weight  be  laid  upon  it,  what  is 
the  pressure  per  sq.  in.  upon  the  upper  head  ? 

SOLUTION.— In  addition  to  the  pressure  of  .426  pound  on  the  area  of 
.04909  sq.  in.,  there  is  now  an  additional  pressure  upon  this  area  of 
5  +  }  =  5. 25  pounds,  and  the  total  pressure  upon  this  area  is  .426  + 

5.25  =  5.676  pounds.     The  pressure  per  square  inch  is  X  5.676  = 

115.6  pounds.     Ans. 

EXAMPLE.— In  Fig.  113,  the  plunger  A,  is  10"  in  diameter,  and  is  forced 
100 lb. 


FIG.  113. 
outwards  by   means  of  a  small  pump  B,  which   supplies  the   press 


MECHANICS.  219 

cylinder  with  water.  The  plunger  <7of  the  pump  7?  is  £'  in  diameter.  If 
a  force  of  100  Ib.  be  applied  to  C  by  means  of  the  lever  D,  how  great  a 
weight  can  the  plunger  A  raise,  if  the  plunger  itself  weighs  400  Ib.  ? 

SOLUTION. — First  find  the  pressure  per  sq.  in.  which  the  plunger  C 
exerts  upon  the  water.  Area  of  C=  (1,)*  X  .7854  =  .19635  sq.  in.  Since 
C  exerts  a  pressure  of  100  Ib.  upon  an  area  of  .19635  sq.  in.,  it  will  exert 

a  pressure  of  100  X  ~^^~  lb-  uPon  an  area  of  1  sq.  in.     This  pressure  is 

transmitted  by  the  water  in  the  tube  E  to  the  cylinder  F,  where  it 
forces  the  plunger  A  upwards  as  the  water  is  forced  into  F.  The  pres- 
sure per  sq.  in.  on  the  bottom  of  A  is  the  same  as  that  exerted  by  C. 

100 

Hence,  the  total  pressure  is  10*  X  .7854  X  -^^f  =  40'000  lb-     This 

. lyooo 

result,  less  the  weight  of  A,  equals  the  load  lifted.  Therefore,  40,000  — 
400  =  39,600  lb.  Ans. 

579.  In  these  examples  on  the  hydraulic  press,  no  allow- 
ance has  been  made  for  the  power  lost  in  overcoming  the 
friction  between  the  cup  leathers  and  the  plunger;  this  varies 
according  to  the  condition  of  the  leathers,  and,  of  course, 
the  smoothness  of  the  plunger;  when  the  leathers  are  in  good 
condition  the  loss  is  about  5$  of  the  total  pressure  on  the 
ram ;  when  the  leathers  are  old,  stiff,  and  dirty  the  loss  may 
amount  to  15^  or  more. 


BUOYANT    EFFECTS   OF    WATER. 

58O.  In  Fig.  114  is  shown  a  6-inch  cube,  entirely  sub- 
merged in  water.  The  lateral  pressures  are  equal  and  in 
opposite  directions.  The  upward  pressure 
acting  on  the  lower  surface  of  the  cube 
=  6  X  6  X  21  X  .03617;  the  downward 
pressure  acting  on  the  top  of  the  cube  = 
6  X  6  X  15  X  .03617,  and  the  difference  = 
6  X  6  X  6  X  .03617  =  the  volume  of  the 
cube  in  cubic  inches  x  the  weight  of  one 
cubic  inch  of  water.  That  is,  the  upward 
pressure  exceeds  the  downward  pressure 
by  the  weight  of  a  volume  of  water  equal 
to  the  volume  of  the  body. 

This  excess  of  upward  pressure  over  the  downward  pressure 


220 


MECHANICS. 


acts  against  gravity ;  that  is,  the  water  presses  the  body 
up  with  a  greater  force  than  it  presses  it  down;  consequently, 
if  u  body  is  immersed  in  a  fluid,  it  will  lose  in  weight  an 
amount  equal  to  the  weight  of  the  fluid  it  displaces.  This  is 
called  the  principle  of  Archimedes,  because  it  was  first 
stated  by  him. 

58 1 .  This  principle  may  be  experimentally  demonstrated 
with  the  beam  scales,  as  shown  in  Fig.  115. 

From  one  scale  pan  suspend  a  hollow  cylinder  of  metal  /, 
and  below  that  a  solid  cylinder  a  of  the  same  size  as  the 

hollow  part  of  the  up- 
per cylinder.  Put 
weights  in  the  other 
scale  pan  until  they  ex- 
actly balance  the  two 
cylinders.  If  a  be  im- 
mersed in  water,  the 
scale  pan  containing  the 
weights  will  descend, 
showing  that  a  has  lost 
some  of  its  weight. 
Now  fill  /  with  water, 
and  the  volume  of  water 
that  can  be  poured  into 
t  will  equal  that  dis- 
placed by  a.  The  scale 
pan  that  contains  the  weights  will  gradually  rise  until  t  is 
filled,  when  the  scales  balance  again. 

If  a  body  be  lighter  than  the  liquid  in  which  it  is  im- 
mersed, the  upward  pressure  will  cause  it  to  rise,  and  project 
partly  out  of  the  liquid,  until  the  weight  of  the  body  and 
the  weight  of  the  liquid  displaced  are  equal.  If  the  im- 
mersed body  be  heavier  than  the  liquid,  the  downward 
pressure  plus  the  weight  of  the  body  will  be  greater  than 
the  upward  pressure,  and  the  body  will  fall  downwards 
until  it  touches  bottom  or  meets  an  obstruction.  If  the 
weights  of  equal  volumes  of  the  liquid  and  the  body  are 


MECHANICS.  221 

equal,  the  body  will  remain  stationary,  and  be  in  equilibrium 
in  any  position  or  depth  beneath  the  surface  of  the  liquid. 

582.  An  interesting  experiment  in  confirmation  of  the 
above  facts  may  be  performed  as  follows:  Drop  an  egg  into  a 
glass  jar  filled  with  fresh  water.  The  mean  density  of  the 
egg  being  a  little  greater  than  that  of  water,  it  will  fall  to 
the  bottom  of  the  jar.  Now,  dissolve  salt  in  the  water, 
stirring  it  so  as  to  mix  the  fresh  and  salt  water.  The  salt 
water  will  presently  become  denser  than  the  egg,  and  the 
egg  will  rise.  Now,  if  fresh  water  be  poured  in  until  the 
egg  and  water  have  the  same  density,  the  egg  will  remain 
stationary  in  any  position  that  it  may  be  placed  below  the 
surface  of  the  water. 


EXAMPLES   FOR   PRACTICE. 

1.  Suppose  a  cylinder  to  be   filled  with  water  and  placed  in  an 
upright  position.     If  the  diameter  of  the  cylinder  be  19",  and  its  total 
length  inside  be  26",  what  will  be  the  total  pressure  on  the  bottom 
when  a  pipe  \"  in  diameter  and  12  ft.  long  is  screwed  into  the  cylinder 
head  and  filled  with  water  ?     The  pipe  is  vertical.  Ans.  1,743.2  Ib. 

2.  In  the  last  example,  what  is  the  total  pressure  against  the  upper 
head?  Ans.  1,476.6  Ib. 

3.  In  example  1,  a  piston  is  fitted  to  the  upper  end  of  the  pipe,  and 
an  additional  force  of  10  Ib.  is  applied  to  the  water  in  the  pipe,    (a) 
What  is  the  total  pressure  on  the  bottom  of  the  cylinder  ?  (t>)  On  the 
upper  head?  j  (a)  16,184  Ib. 

IS-  1  (b)  15.906  Ib. 

4.  In  example  3,  what  is  the  pressure  per  square  inch  in  the  pipe  2* 
from  the  upper  cylinder  head  ?  Ans.  56.0656  Ib.  per  sq.  in. 

5.  A  water  tower  80  feet  high  is  filled  with  water.     A  pipe  4"  in 
diameter  is  so  connected  to  the  side  of  the  tower  that  its  center  is 
3  feet  from  the  bottom.     If  the  pipe  is  closed  by  a  flat  cover,  what  is  the 
total  pressure  against  the  cover  ?  Ans.  420  Ib. 

6.  In  the  last  example,  what  is  the  upward  pressure  per  square 
inch  10  feet  from  the  bottom  of  the  tower  ? 

Ans.  30.3828  Ib.  per  sq.  in. 

7.  A  cube  of  wood,  one  edge  of  which  measures  3  feet,  is  sunk  until 
the  upper  surface  is  40  feet  below  the  level  of  the  water;  what  is  the 
total  force  which  tends  to  move  the  cube  upwards  ?  Ans.  1,687.5  Ib. 


222 


MECHANICS. 


PNEUMATICS. 

583.     Pneumatics  is  that  branch  of  mechanics  which 
treats  of  the  properties  of  gases. 


584.  The  most  striking  feature  of  all  gases  is  their  ex- 
treme expansibility.    If  we  inject  a  portion  of  gas,  however 

small  it  may  be,  into  a  vessel, 
it  will  expand  and  fill  that 
vessel.  If  a  bladder  or  foot- 
ball be  partly  filled  with  air, 
and  placed  under  a  glass  jar 
(called  a  receiver),  from  which 
the  air  has  been  exhausted, 
the  bladder  or  football  will 
immediately  expand,  as  shown 
in  Fig.  116.  The  force  which 
a  gas  always  exerts  when  con- 
fined in  a  limited  space  is 
called  tension.  The  word 
tension  in  this  case  means 

pressure,  and   is   only  used   in   this  sense  in  reference  to 

gases. 

585.  As  water  is  the  most  common  type  of  fluids  so  air 
is  the  most  common  type  of  gases.      It  was  supposed  by  the 
ancients  that  air  had  no  weight,  and  it  was  not  until  about 
the  year  1650  that  it  was  proven  that  air  really  had  weight. 
A   cubic   inch    of   air,    under   ordinary   conditions,    weighs 
.31  grain,  nearly.     At  a  temperature  of  32°  and  a  pressure  of 
14.7  pounds  per  square   inch,  the  ratio  of  the  weight  of 
air  to  water  is  about  1  :  774;  that  is,  air  is   only   ^  as 
heavy  as  water.     In  Art.  58O,  it  was  shown  that  if  a  body 
were  immersed  in  water,  and  weighed  less  than  the  volume 
of  water  displaced,  the  body  would  rise  and  project  partly 
out  of  the  water.     The  same  is  true  to  a  certain  extent  of 
air.      If  a  vessel  made  of  light  material  be  filled  with  a  gas 
lighter  than  air,  so  that  the  total  weight  of  the  vessel  and 


MECHANICS. 


gas  is  less  than  the  air  they  displace,  the  vessel  will  rise, 
is  on  this  principle  that  balloons  are  made. 


It 


586.  Since  air  has  weight,  it  is  evident  that  the  enor- 
mous quantity  of  air  that  constitutes  the  atmosphere  must 
exert  a  considerable  pressure  upon  the  earth.  This  is  easily 
proven  by  taking  a  long  glass  tube  closed  at  one  end,  and 
filling  it  with  mercury.  If  the  finger  be  placed  over  the  open 
end  so  as  to  keep  the  mercury  from  running  out,  and  the 
tube  inverted  and  placed  in  a  cup  of  mercury,  as  shown  in 
Fig.  117.  the  mercury  will  fall,  then  rise,  and  after  a  few 
oscillations  will  come  to  rest  at  a  height  above  the  top  of 
the  mercury  in  the  glass  equal  to  about  30  inches.  This 
height  will  always  be  the  same  under 
the  same  atmospheric  conditions. 
Now,  if  the  atmosphere  has  weight, 
it  must  press  upon  the  upper  sur- 
face of  the  mercury  in  the  glass  with 
equal  intensity  upon  every  square 
unit,  except  upon  that  part  of  the 
surface  occupied  by  the  tube.  In 
order  that  there  may  be  equilibrium, 
the  weight  of  the  mercury  in  the  tube 
must  be  equal  to  the  pressure  of  the 
air  upon  a  portion  of  the  surface  of  the 
mercury  in  the  glass,  equal  in  area 
to  the  inside  of  the  tube.  Suppose 
that  the  area  of  the  inside  of  the  tube 
is  1  square  inch,  then,  since  mercury 
is  13.6  times  as  heavy  as  water, 
the  weight  of  the  mercury  column 
is  .03617  X  13.6  X  30  =  14.7574  Ib. 
The  actual  height  of  the  mercury  is 
a  little  less  than  30  inches,  and  the 
actual  weight  of  a  cubic  inch  of  dis- 
tilled water  is  a  little  less  than  .03617 
Ib.  When  these  considerations  are 
taken  into  account,  the  average  weight  of  the  mercurial 


224  MECHANICS. 

column  at  the  level  of  the  sea,  when  the  temperature  is  60°, 
is  14.69  lb.,  or  practically  14.7  Ib.  Since  this  weight,  when 
exerted  upon  1  square  inch  of  the  liquid  in  the  glass,  just 
produces  equilibrium,  it  is  plain  that  the  pressure  of  the 
outside  air  is  14.7  lb.  upon  every  square  inch  of  surface. 

587.  Vacuum. — The  space  between  the  upper  end  of 
the  tube  and  the  upper  surface  of  the  mercury  is  called  a 
vacuum,  meaning  that  it  is  an  entirely  empty  space,  and 
does  not  contain  any  substance,  solid,  liquid,   or  gaseous. 
If  there  were  a  gas  of  some  kind  there,  no  matter  how  small 
the  quantity  might  be,  it  would  expand,  filling  the  space, 
and  its  tension  would  cause  the  column  of  mercury  to  fall 
and  become  shorter,  according  to  the  amount  of  gas  or  air 
present.   The  space  is  then  called  a  partial  vacuum.  If  the 
mercury  fell   1  inch,  so  that  the  column  was  only  29  inches 
high,  we  should  say,  in  ordinary  language,  that  there  were 
29  inches  of  vacuum.     If  it  fell  8  inches,  we  should  say  that 
there  were  22  inches  of  vacuum ;  if  it  fell  16  inches,  we  should 
say  that  there  were  14  inches  of  vacuum,  and  so  on.     Hence, 
when   the  vacuum    gauge    of   a   condensing   engine    shows 
26  inches  of  vacuum,  there  is  enough  air  in  the  condenser  to 

produce  a  pressure  of  —         —  X  14.7  =  —  X  14.7  =  1.96  lb. 

O\J  -  o(j 

per  sq.  in.  In  all  cases  where  the  mercury  column  is  used  to 
measure  a  vacuum,  the  height  of  the  column  in  inches  gives 
the  number  of  inches  of  vacuum.  Were  the  column  only  5* 
high,  the  vacuum  would  be  5". 

If  the  tube  had  been  filled  with  water  instead  of  mercury, 
the  height  of  the  column  of  water  to  balance  the  pressure 
of  the  atmosphere  would  have  been  30  X  13.6  =  408  inches 
=  34  feet.  This  means  that,  if  a  tube  be  filled  with  water, 
inverted,  and  placed  in  a  dish  of  water  in  a  manner  similar 
to  the  experiment  made  with  the  mercury,  the  height  of  the 
column  of  water  will  be  34  feet. 

588.  The  barometer  is  an  instrument  used  for  meas- 
uring the  pressure  of  the  atmosphere.     There  are  two  kinds 


MECHANICS. 


225 


in   general  use— the   mercurial  barometer  and  the  aneroid 
barometer.      The     mercurial     barometer     is 

shown  in  Fig.  118.  The  principle  is  the  same 
as  the  inverted  tube  shown  in  Fig.  117.  In  this 
case  the  tube  and  the  cup  at  the  bottom  are  pro- 

Ii^pta  tected  by  a  brass  or  iron  casing.  At  the  top  of 
j!BJt  the  tube  is  a  graduated  scale  which  can  be  read  to 
jiffij  TTnru  °f  an  mcn  by  means  of  a  vernier.  Attached 
to  the  casing  is  an  accurate  thermometer  for 
J:H^  determining  the  temperature  of  the  outside  air  at 
the  time  the  barometric  observation  is  taken. 
This  is  necessary,  since  mercury  expands  when 
the  temperature  is  increased,  and  contracts  when 
the  temperature  falls;  for  this  reason  a  standard 
temperature  is  assumed,  and  all  barometer  read- 
ings are  reduced  to  this  temperature.  This  stand- 
ard temperature  is  usually  taken  at  32°  F.,  at 
which  temperature  the  height  of  the  mercurial 
column  is  30  inches.  Another  correction  is  made 
for  the  altitude  of  the  place  above  sea-level,  and 
a  third  correction  for  the  effects  of  capillary 
attraction. 

589.  In  Fig.  119  is  an  illustration  of  an 
aneroid  barometer.  These  instruments  are 
made  in  various  sizes,  from  the  size  of  a  watch  up 
to  8  or  10  inches  in  diameter.  They  consist  of  a 
cylindrical  box  of  metal  with  a  top  of  thin,  elastic 
corrugated  metal.  The  air  is  exhausted  from  the 
box.  When  the  atmospheric  pressure  increases, 
the  top  is  pressed  inwards,  and  when  it  is  dimin- 
'FIG.  us.  ished,  the  top  is  pressed  outwards  by  its  own  elas- 
ticity, aided  by  a  spring  beneath.  These  movements  of  the 
cover  are  transmitted  and  multiplied  by  a  combination  of 
delicate  levers,  which  act  upon  an  index  hand,  and  cause  it 
to  move  either  to  the  right  or  left,  over  a  graduated  scale. 
These  barometers  are  self-correcting  (compensated)  for  vari- 
ations in  temperature.  They  are  very  portable,  occupying 


226  MECHANICS. 

but  a  small  space,  and  are  so  delicate  that  they  are  said  to 
show  a  difference  in  the  atmospheric  pressure  when  trans- 


ferred from  the  table  to  the  floor.     The  mercurial  barometer 
is  the  standard. 

59O.  With  air,  as  with  water,  the  lower  we  get  the 
greater  the  pressure,  and  the  higher  we  get  the  less  the 
pressure.  At  the  level  of  the  sea,  the  height  of  the  mercurial 
column  is  about  30  inches;  at  5, 000  feet  above  the  sea,  it  is 
24.7  inches;  at  10,000  feet  above  the  sea,  it  is  20. 5  inches-, 
at  15,000  feet,  it  is  16.9  inches;  at  3  miles,  it  is  16.4  inches, 
and  at  6  miles  above  the  sea -level,  it  is  8.9  inches. 

The  density  or  weight  of  the  atmosphere  also  varies  with 
the  altitude;  that  is,  a  cubic  foot  of  air  at  an  elevation  of 
5,000  feet  above  the  sea-level  will  not  weigh  as  much  as  a 
cubic  foot  at  sea-level.  This  is  proven  conclusively  by  the 


MECHANICS.  227 

fact  that  at  a  height  of  3|  miles  the  mercurial  column 
measures  but  15  inches,  indicating  that  half  the  weight  of 
the  entire  atmosphere  is  below  that.  It  is  known  that  the 
height  of  the  earth's  atmosphere  is  at  least  50  miles;  hence, 
the  air  just  before  reaching  the  limit  must  be  in  an  exceed- 
ingly rarefied  state.  It  is  by  means  of  barometers  that 
great  heights  are  measured.  The  aneroid  barometer  has 
the  heights  marked  on  the  dial,  so  that  they  can  be  read 
directly.  With  the  mercurial  barometer,  the  heights  must 
be  calculated  from  the  reading. 

591.  The  atmospheric  pressure  is  everywhere  present, 
and  presses  all  objects  in  all  directions  with  equal  force.  If 
a  book  is  laid  upon  the  table,  the  air  presses  upon  it  in  every 
direction  with  an  equal  average  force  of  14.7  pounds  per 
square  inch.  It  would  seem  as  though  it  would  take  con- 
siderable force  to  raise  a  book  from  the  table,  since,  if  the 
size  of  the  book  were  8  inches  by  5  inches,  the  pressure  upon 
it  would  be  8  X  5  X  14.7  =  588  Ib.  ;  but  there  is  an  equal 
pressure  beneath  the  book  which  counteracts  the  pressure 
on  the  top.  It  would  now  seem  as  though  it  would  require 
a  great  force  to  open  the  book,  since  there  are  two  pressures 
of  588  pounds  each  acting  in  opposite  directions  and  tend- 
ing to  crush  the  book-,  so  it  would,  but  for  the  fact  that 
there  is  a  layer  of  air  between  each  leaf  acting  upwards  and 
downwards  with  a  pressure  of  14.7  pounds  per  square  inch. 
If  two  metal  plates  be  made  as  perfectly  smooth  and  flat  as 
it  is  possible  to  get  them,  and  the  edge  of  one  be  laid  upon 
the  edge  of  the  other,  so  that  one  may  be  slid  upon  the 
other  and  thus  exclude  the  air,  it  will  take  an  immense 
force,  compared  to  the  weights  of  the  plates,  to  separate 
them.  This  is  because  the  full  pressure  of  14  7  pounds  per. 
square  inch  is  then  exerted  upon  each  plate,  with  no 
counteracting  equal  pressure  between  them. 

If  a  piece  of  flat  glass  be  laid  upon  a  flat  surface  that  has 
been  previously  moistened  with  water,  it  will  require  con- 
siderable force  to  separate  them ;  this  is  because  the  water 
helps  to  fill  up  the  pores  in  the  flat  surface  and  glass,  and 


228  MECHANICS. 

thus  creates  a  partial  vacuum  between  the  glass  and  the  sur- 
face, thereby  reducing  the  counter  pressure  beneath  the  glass. 

592.  Tension  of  Gases. — In  Fig.  117,  the  space  above 
the  column  of  mercury  was  said  to  be  a  vacuum,  and  it  was 
also  said  that  if  any  gas  or  air  were  present,  it  would  ex- 
pand, its  tension  forcing  the  column  of  mercury  downwards. 
If  sufficient  gas  be  admitted  to  cause  the  mercury  to  stand 

14.7 
at  15  inches,  the  tension  of  the  gas  is  evidently  — ^—  =  7.35  Ib. 

per  square  inch,  since  the  pressure  of  the  outside  air,  14  7  Ib. 
per  square  inch,  now  balances  only  15  instead  of  30  inches 
of  mercury;  that  is,  it  balances  only  half  as  much  as  it 
would  if  there  were  no  gas  in  the  tube;  therefore,  the  pres- 
sure (tension)  of  the  gas  in  the  tube  is  7.35  pounds  If  more 
gas  be  forced  into  the  tube  until  the  top  of  the  mercurial 
column  is  just  level  with  the  mercury  in  the  cup,  the  gas  in 
the  tube  will  then  have  a  tension  equal  to  the  outside  pres- 
sure of  the  atmosphere.  Suppose  that  the  bottom  of  the 
tube  is  fitted  with  a  piston,  and  that  the  total  length  of  the 
inside  of  the  tube  is  36  inches.  If  the  piston  be  shoved  up- 
wards so  that  the  space  occupied  by  the  gas  is  18  inches 
long  instead  of  36  inches,  the  temperature  remaining  the 
same  as  before,  it  will  be  found  that  the  tension  of  the  gas 
within  the  tube  is  29.4  Ib.  It  will  be  noticed  that  the 
volume  occupied  by  the  gas  is  only  half  that  in  the  tube 
before  the  piston  was  moved,  while  the  pressure  is  twice  as 
great,  since  14.7  X  2  —  29.4  Ib.  If  the  piston  be  shoved  up, 
so  that  the  space  occupied  by  the  gas  is  only  9  inches,  in- 
stead of  18  inches,  the  temperature  still  remaining  the 
same,  the  pressure  will  be  found  to  be  58.8  pounds  per 
square  inch.  The  volume  has  again  been  reduced  one-half, 
and  the  pressure  increased  two-fold,  since  29.4  X  2  =  58.8  Ib. 
The  space  now  occupied  by  the  gas  is  9  inches  long,  whereas, 
before  the  piston  was  moved,  it  was  36  inches  long;  as  the 

tube  was  assumed  to  be  of  uniform  diameter  throughout  its 
q         -i 

'length,  the  volume  is  now  —  =  —  of  its  original  volume,  and 
ob       4 


MECHANICS.  229 

58  8 

its  pressure  is—  '—=  4  times  its  original  pressure.  More- 
over, if  the  temperature  of  the  confined  gas  remains  the 
same,  the  pressure  and  volume  will  always  vary  in  a  similar 
way. 

593.  The  law  which  states  these  effects  is  called 
Mariotte's  law,  and  is  as  follows : 

Mariotte's  Law. —  The  temperature  remaining  the  same, 
the  volume  of  a  given  quantity  of  gas  varies  inversely  as  the 
pressure. 

The  meaning  of  the  law  is  this:  If  the  volume  of  a  gas 
be  diminished  to  £,  £,  ^,  etc.,  of  its  former  volume,  the  ten- 
sion will  be  increased  2,  3,  5,  etc.,  times,  or  if  the  outside 
pressure  be  increased  2,  3,  5,  etc.,  times,  the  volume  of  the 
gas  will  be  diminished  to  £,  £,  \,  etc.,  of  its  original  volume, 
the  temperature  remaining  constant.  It  also  means  that  if 
a  gas  is  under  a  certain  pressure,  and  this  pressure  is  dimin- 
ished to  £,  £,  jV,  etc.,  of  its  original  intensity,  the  volume  of 
the  confined  gas  will  be  increased  2,  4,  10,  etc.,  times — its 
tension  decreasing  at  the  same  rate. 

Suppose  3  cubic  feet  of  air  to  be  under  a  pressure  of 
60  pounds  per  square  inch  in  a  cylinder  fitted  with  a  movable 
piston,  then  the  product  of  the  volume  and  pressure  is 
3  x  60  =  180.  Let  the  volume  be  increased  to  6  cubic  feet, 
then  the  pressure  will  be  30  pounds  per  square  inch,  and 
30  x  6  =  180,  as  before.  Let  the  volume  be  increased  to 

24 

24  cubic  feet;  it  is  then  ^—  =  8  times  its  original  volume, 
o 

and  the  pressure  is  £  of  its  original  pressure,  or  GO  X  |  =  7£ 
lb.,  and  24  x  7|  =  180,  as  in  the  two  preceding  cases.  It 
will  now  be  noticed  that  if  a  gas  be  enclosed  within  a  con- 
fined space,  and  allowed  to  expand  without  losing  any  heat, 
t  lie  product  of  the  pressure  and  the  corresponding  volume  for 
any  one  position  of  the  piston  is  the  same  as  for  any  other 
position.  If  the  piston  were  forced  inwards  so  as  to  com- 
press the  air,  the  same  results  would  be  obtained. 


230  MECHANICS. 

594.  If  the  volume  of  the  vessel  and  the  pressure  of 
the  gas  are  known,  and  it  is  desired  to  know  the  pressure 
after  the  first  volume  has  been  changed : 

Rule  1O9. — Divide  the  product  of  the  first,  or  original, 
-volume  and  pressure  by  the  new  volu'me;  the  result  will  be  the 
new  pressure. 

Or,  let  /  =  original  pressure ; 
/,  =  final  pressure ; 

v  =  volume  corresponding  to  the  pressure  />; 
i>^  =  volume  corresponding  to  the  pressure  /,. 

Then,  A  =  e 

EXAMPLE. — At  the  point  of  cut-off  in  a  steam  engine,  the  amount  of 
steam  in  the  cylinder  is  862  cu,  in.  The  pressure  at  this  point  is  120 
Ib.  per  sq.  in.  What  will  be  the  pressure  of  the  steam  when  the  piston 
has  reached  the  end  of  its  stroke,  and  the  volume  is  1,800  cu.  in.  ? 

SOLUTION. — Applying  the  rule,  —         —  =  57.47  Ib.  per  sq.  in.    Ans. 

595.  If  it  is  required  to  determine  the  volume  after  a 
change  in  the  pressure: 

Rule  11O. — Divide  the  product  of  the  original  volume 
and  pressure  by  the  new  pressure  ;  the  result  will  be  the  new 
volume. 

Or,  using  the  same  letters  as  before, 

*<=£•       .  .    • 

EXAMPLE. — At  the  commencement  of  compression,  the  volume  of 
the  steam  is  380  cu.  in.,  and  the  pressure  is  18  Ib.  per  sq.  in.  At  the 
end  of  compression,  the  pressure  is  112  Ib.  per  sq.  in.  What  is  the  final 
volume  ? 

OQA  -y   1  Q 

SOLUTION.— Applying  the  rule,  — ^ —  =  61.07  cu.  in.     Ans. 

EXAMPLE.— A  vessel  contains  10  cu.  ft.  of  air  at  a  pressure  of  15  Ib, 
per  sq.  in.,  and  has  25  cu.  ft.  of  air  of  the  same  pressure  forced  into  it; 
what  is  the  resulting  pressure  ? 

SOLUTION.  —The  original  volume  =  10  +  25  =  35  cu.  ft.  The  original 
pressure  is  15  Ib.  per  sq.  in.  The  final  volume  is  10  cu.  ft.  Hence, 

applying  rule  1O9,  — wf—  =  52/5  Ib.  per  sq.  in.     Ans. 


MECHANICS. 


231 


It  must  be  remembered  that  in  the  preceding  examples 
the  temperature  is  supposed  to  remain  constant. 


EXAMPLES    FOR    PRACTICE. 

1.  A  vessel  contains  25  cubic  feet  of  gas,  at  a  pressure  of  18  Ib.  per 
sq.  in. ;  if  125  cu.  ft.  of  gas  having  the  same  pressure  are  forced  into 
the  vessel,  what  will  be  the  resulting  pressure  ?    Ans.  108  Ib.  per  sq.  in. 

2.  The  volume  of  steam  in  the  cylinder  of  a  steam  engine  at  cut-off 
is  1.35  cu.  ft.,  and  the  pressure  is  85  Ib.  per  sq.  in.     The  pressure  at  the 
end  of  the  stroke  is  25  Ib.  per  sq.  in.     What  is  the  new  volume  ? 

Ans.  4.59  cu.  ft. 

3.  A  receiver  contains  180  cu.  ft.  of  gas,  at  a  pressure  of  20  Ib.  per 
sq.  in. ;   if  a  vessel  holding  12  cu.  ft.  be  filled  from  the  larger  vessel 
until  its  pressure  is  20  Ib.  per  sq.  in.,  what  will  be  the  pressure  in  the 
larger  vessel  ?  Ans.  18J  Ib.  per  sq.  in. 

4.  A  spherical  shell  has  a  part  of  the  air  within  it  removed,  forming 
a  partial  vacuum ;  -if  the  outside  diameter  of  the  shell  is  18",  and  the 
pressure  of  the  air  within  is  5  Ib.  per  sq.  in.,  what  is  the  total  pressure 
tending  to  crush  the  shell  ?  Ans.  9,873.42  Ib. 

PNEUMATIC    MACHINES. 

596.     The  Air  Pump. —  The  air  pump  is  an  instrument 
for  removing  air  from  a  given  space.     A   section   of  the 


principal  parts  is  shown  in  Fig.  120,  and  the  complete  instru- 
ment in  Fig.  121.  The  closed  vessel  R  is  called  the 
receiver,  and  the  space  which  it  encloses  is  that  from  which 
it  is  desired  to  remove  the  air.  It  is  usually  made  of  glass, 


232 


MECHANICS. 


and  the  edges  are  ground  so  as  to  be  perfectly  air-tight. 
When  made  in  the  form  shown,  it  is  called  a  bell-jar 
receiver.  The  receiver  rests  upon  a  horizontal  plate,  in  the 
center  of  which  is  an  opening  communicating  with  the  pump 
cylinder  C  by  means  of  the  passage  /  /.  The  pump  piston 
fits  the  cylinder  accurately,  and  has  a  valve  V  opening  up- 
wards. Where  the  passage  /  /  joins  the  cylinder,  is  another 
valve  F,  also  opening  upwards.  When  the  piston  is  raised, 
the  valve  V  closes,  and,  since  no  air  can  get  into  the  cylin- 
der from  above,  the  piston  leaves  a  vacuum  behind  it.  The 
pressure  upon  V  being  now  removed,  the  tension  of  the  air 
in  the  receiver  R  causes  V  to  rise;  the  air  in  the  receiver 
and  passage  t  t  then  expands  so  as  to  occupy  the  additional 
space  provided  by  the  upward  movement  of  the  piston. 

The  piston  is  now 
pushed  down,  the 
valve  V  closes,  the 
valve  V  opens,  and 
the  air  in  C  escapes. 
The  lower  valve  V  is 
sometimes  supported, 
as  shown  in  Fig.  120, 
by  a  metal  rod  passing 
through  the  piston, 
and  fitting  it  some- 
what tightly.  When 
the  piston  is  raised  or 
lowered,  this  rod 
moves  with  it.  A  but- 
ton near  the  upper  end 
of  the  rod  confines  its 
motion  within  very 
narrow  limits,  the  pis- 
ton sliding  upon  the 
rod  during  the  greater  part  of  the  journey. 

In  the  complete  form  of  the  instrument  shown  in  Fig.  121, 
communication  between  receiver  and  pump  is  made  by 
means  of  the  tube  t. 


MECHANICS. 


233 


597.  Degrees  and  Limits  of  Exhaustion. — Sup- 
pose that  the  volume  of  R  and  /  together  is  four  times  that 
of  C,  Fig.  121,  and  that  there  are  say  200  grains  of  air 
in  R  and  /,  and  50  grains  in  C  when  the  piston  is  at 
the  top  of  the  cylinder.  At  the  end  of  the  first  stroke, 
when  the  piston  is  again  at  the  top,  50  grains  of  air 
in  the  cylinder  C  will  have  been  removed,  and  the 
200  grains  in  R  and  t  will  occupy  the  space  R,  t,  and  C.  The 
ratio  between  the  sum  of  the  spaces  R  and  /,  and  the 
total  space  R  -\-  t  +  C  is  | ;  hence,  200  X  -|  —  100  grains  = 
the  weight  of  air  in  R  and  t  after  the  first  stroke.  After  the 
second  stroke,  the  weight  of  the  air  in  R  and  /  would  be 
(200  X  |)  X  i  =-200  X  (-1)'  =  200  X  If  =  128  grains.  At 
the  end  of  the  third  stroke  the  weight  would  be  [200  X  (-J)*] 
X  |  =  200  X  (|)3  =  200  X  TQ2T=  103-4  grains.  At  the  end 
of  »  strokes  the  weight  would  be  200  X  (f )".  It  is  evident 
that  //  is  impossible  to  remove  all  of  the  air  tJiat  is  contained 
in  R  and  t  by  this  method.  It  requires  an 
exceedingly  good  air  pump  to  reduce  the 
tension  of  the  air  in  R  to  -fa  of  an  inch  of 
mercury.  When  the  air  has  become  so 
rarefied  as  this,  the  valve  V  will  not  lift, 
and,  consequently,  no  more  air  can  be 
exhausted. 


598.      Magdeburg    Hemispheres. 

— By  means  of  the  two  hemispheres  shown 

in  Fig.   122,   it   can  be   proven  that   the 

atmosphere  presses  upon  a  body  equally 

in   all  directions.      They   were   invented 

by   Otto   Von  Guericke,   of   Magdeburg, 

and  are  called  the  Magdeburg  hemispheres. 

One  of  the  hemispheres  is  provided  with  a 

stop-cock,  by  which  it  can  be  screwed  on 

to  an  air  pump.     The  edges  fit  accurately,  and   are  well 

greased,  so  as  to  be  air-tight.     As  long  as  the  hemispheres 

contain  air,  they  can  be  separated  with  no  difficulty;  but 

when  the  air  in  the  interior  is  pumped  out  by  means  of  an 


234 


MECHANICS. 


air  pump,  they  can  be  separated  only  with  great  difficulty. 
The  force  required  to  separate  them  will  be  equal  to  the 
area  of  the  largest  circle  of  the  hemisphere  in  square  inches 
multiplied  by  14.7  pounds. 

This  force  will  be  the  same  in  whatever  position  the  hem- 
isphere may  be  held,  thus  proving  that  the  pressure  of  air 
upon  it  is  the  same  in  all  directions. 

NOTE.— A  theoretically  perfect  vacuum  is  sometimes  called  a 
Torricellian  vacuum. 

599.  The  Weight  Lifter. — The  pressure  of  the  at- 
mosphere is  very  clearly  shown  by  means  of  an  apparatus 
like  that  illustrated  in  Fig.  123.  Here  a 
cylinder  fitted  with  a  piston  is  held  in  sus- 
pension by  a  chain.  At  the  top  of  the 
cylinder  is  a  plug  A  which  can  be  taken 
out.  This  plug  is  removed,  the  piston 
pushed  up  (the  force  necessary  being  equal 
to  the  weight  of  the  piston  and  rod  B) 
until  it  touches  the  cylinder  head.  The 
plug  is  then  screwed  in,  and  the  piston 
will  remain  at  the  top  until  a  weight  has 
been  hung  on  the  rod  equal  to  the  number 
of  pounds  obtained  by  multiplying  the 
number  of  square  inches  in  the  piston  by 
14.7  pounds,  and  subtracting  therefrom 
the  weight  of  the  piston  and  rod  in  pounds. 
If  a  force  sufficiently  great  were  em- 
ployed to  pull  the  piston  downwards,  and 
then  any  less  weight  were  attached,  as 
shown  in  Fig.  123,  the  piston  would  ascend 
and  carry  the  weight  up  with  it. 

6OO.     Suppose  the  weight   to  be    re- 
moved, and  the  piston  to  be  supported,  say 
midway   of   the   length    of   the   cylinder. 
Let  the  plug  be  removed,  and  air  admit- 
FIG.  123.  ted  above  the  piston,  then  screw  the  plug 

back    into   its    place;    if   the  piston    be    shoved    upwards, 


MECHANICS  235 

the  further  up  it  goes  the  greater  will  be  the  force  neces- 
sary to  push  it,  on  account  of  the  compression  of  the 
air.  If  the  piston  is  of  large  diameter  it  will  also  require 
a  great  force  to  pull  it  out  of  the  cylinder,  as  a  little  consid- 
eration will  show.  For  example,  let  the  diameter  of  the 
piston  be  20  inches,  the  length  of  the  cylinder  36  inches,  and 
the  weight  of  the  piston  and  rod  100  pounds.  If  the  piston 
is  in  the  middle  of  the  cylinder,  there  will  be  18  inches  of 
space  above  it,  and  18  inches  of  space  below  it.  The  area  of 
the  piston  is  20s  X  .7854  =  314.16  square  inches,  and  the  at- 
mospheric pressure  upon  it  is  314.16X14.7  =  4,618  lb., 
nearly.  In  order  to  shove  the  piston  upwards  9  inches,  the 
pressure  upon  it  must  be  twice  as  great,  or  9,236  pounds, 
and  to  this  must  be  added  the  weight  of  the  piston  and  rod, 
or  9,236  -f  100  =  9,336  lb.  The  force  necessary  to  cause  the 
piston  to  move  upwards  9  inches  would  then  be  9,336  —  4,618 
=  4,718  lb.  Now,  suppose  the  piston  to  be  moved  down- 
wards until  it  is  just  on  the  point  of  being  pulled  out  of  the 
cylinder.  The  volume  above  it  will  then  be  twice  as  great 
as  before,  and  the  pressure  one-half  as  great,  or  4,618  -f-  2 
=  2,309  lb.  The  total  upward  pressure  will  be  the  pressure 
of  the  atmosphere  less  the  weight  of  the  piston  and  rod,  or 
4,618  —  100  =  4,518  lb.,  and  the  force  necessary  to  pull  it 
downwards  to  this  point  will  be  4,518  —  2,309  =  2,209  lb. 

'6O1.  Air  Compressor*. — For  many  purposes  com- 
pressed air  is  preferable  to  steam  or  other  gas  for  use  as  a 
motive  power.  In  such  cases  air  compressors  are  used 
to  compress  the  air.  These  are  made  in  many  forms,  but 
the  most  common  one  is  to  place  a  cylinder,  called  the  air 
cylinder,  in  front  of  the  cross-head  of  a  steam  engine,  so 
that  the  piston  of  the  air  cylinder  can  be  driven  by  at- 
taching its  piston  rod  to  the  cross-head,  in  a  manner  similar 
to  a  steam  pump.  A  cross-section  of  the  air  cylinder  of  a 
compressor  of  this  kind  is  shown  in  Fig.  124,  in  which  A  is 
the  piston,  and  B  is  the  piston  rod,  driven  by  the  cross-head 
of  a  steam  engine,  not  shown  in  the  figure.  Both  ends  of 
the  lower  half  of  the  cylinder  are  fitted  with  inlet  valves  D 


236 


MECHANICS 


and  D'  which  allow  the  air  to  enter  the  cylinder,  and  both 
ends  of  the  upper  half  are  fitted  with  discharge  valves  F  and 


FIG.  124. 

F'  which  allow  the  air  to  escape  from  the  cylinder  after  it 
has  been  compressed  to  the  required  pressure. 

6O2.  Suppose  the  piston  A  to  be  moving  in  the  direction 
of  the  arrow,  then  all  the  inlet  valves  D  in  the  lower  half  of  the 
left-hand  end  of  the  cylinder  from  which  the  piston  is  moving 
will  be  forced  inwards  by  the  pressure  of  the  atmosphere, 
which  overcomes  the  resistance  of  the  spring  C  tending  to 
keep  the  valve  on  its  seat,  thus  allowing  the  air  to  rush  into 
the  cylinder.  On  the  other  side  of  the  piston  the  air  is  being 
compressed,  and,  consequently,  it  forces  the  inlet  valves  D' 
in  the  lower  half  of  the  right-hand  end  of  the  cylinder  to  their 
seats  against  the  resistance  of  their  springs  S.  In  the  upper 
half  of  the  right-hand  end  of  the  cylinder  the  discharge 


MECHANICS. 


237 


valves  P  are  opened  by  the  compressed  air  against  the  resist- 
ance of  the  springs  £',  and  in  the  upper  half  of  the  left-hand 
end  of  the  cylinder  the  discharge  valves  F  are  pressed 
against  their  seats  by  the  springs  E,  the  tension  of  these 
springs  being  so  adjusted  that  the  valves  can  not  be  forced 
from  their  seats  until  the  air  on  the 
other  side  has  been  compressed  to 
the  required  pressure  per  square 
inch.  For  example,  suppose  it  is  de- 
sired to  compress  the  air  to  59  pounds 
per  square  inch,  and  we  wish  to  find 
at  what  point  of  the  stroke  the  dis- 
charge valves  will  open,  they  having 
been  set  for  this  pressure.  Now, 
59  pounds  per  square  inch  =  4  atmos- 
pheres very  nearly;  hence,  the  vol- 
ume must  be  £  of  the  volume  at  the 
beginning  of  the  stroke,  or  the  valves 
will  open  when  the  piston  has  trav- 
eled £  of  its  stroke. 

The  air,  after  being  discharged 
from  the  cylinder,  passes  out  through 
the  discharge  pipe  H,  and  from 
thence  is  conveyed  to  its  destination. 

6O3.     Hero's    Fountain. — 

Hero's    fountain    derives   its   name 
from  its  inventor,    Hero,  who 
lived  at  Alexandria  120  B.  C. ;  it 
is  shown  in  Fig.  125.   It  depends 
for  its  operation  upon  the  elastic  FIG   125- 

properties  of  air.  It  consists  of  a  brass  dish  A,  and  two  glass 
globes  B  and  C.  The  dish  communicates  with  the  lower  part 
of  the  globe  C  by  means  of  a  long  tube  D,  and  another  tube  E 
connects  the  two  globes.  A  third  tube  passes  through  the 
dish  A  to  the  lower  part  of  the  globe  B.  This  last  tube  being 
taken  out  thegiobe  B  is  partially  filled  with  water,  the  tube 
is  then  replaced,  and  water  is  poured  into  the  dish.  The 


238 


MECHANICS. 


water  flows  through  the  tube  D  into  the  lower  globe,  and  ex- 
pels the  air,  which  is  forced  into  the  upper  globe.  The  air 
thus  compressed  acts  upon  the  water  and  makes  it  jet  out  as 
represented  in  the  figure.  "  Were  it  not  for  the  resistance  of 
the  atmosphere  and  friction,  the  water  would  rise  to  a  height 
above  the  water  in  the  dish  equal  to  the  difference  of  the 
level  of  the  water  in  the  two  globes. 

6O4.     The  Siphon. — The  action  of  the  siphon  illustrates 
the  effect  of  atmospheric  pressure.      It  is  simply  a  bent  tube 

of  unequal  branches,  open  at 
both  ends,  and  is  used  to  con- 
vey a  liquid  from  a  higher 
point  to  a  lower,  over  an  in- 
termediate point  higher  than 
either.  In  Fig.  126,  A  and  B 
are  two  vessels,  B  being  lower 
than  A,  and  A  C B  is  the  bent 
tube,  or  siphon.  Suppose  this 
tube  to  be  filled  with  water 
and  placed  in  the  vessels  as' 
shown,  with  the  short  branch 
A  C  in  the  vessel  A.  The 
water  will  flow  from  the  vessel 
A  into  B,  as  long  as  the  level 
of  the  water  in  B  is  below  the 
FIG.  120.  level  of  the  water  in  A,  and 

the  level  of  the  water  in  A  is  above  the  lower  end  of  the  tube 
A  C.  The  atmospheric  pressure  upon  the  surfaces  of 
A  and  B  tends  to  force  the  water  up  the  tubes  A  C 
and  B  C.  When  the  siphon  is  filled  with  water  each  of 
these  pressures  is  counteracted  in  part  by  the  pressure 
of  the  water  in  that  branch  of  the  siphon  which  is  im- 
mersed in  the  water  upon  which  the  pressure  is  exerted. 
The  atmospheric  pressure  opposed  to  the  weight  of  the 
longer  column  of  water  will,  therefore,  be  more  resisted 
than  that  opposed  to  the  weight  of  the  shorter  column; 
consequently,  the  pressure  exerted  upon  the  shorter  column 


MECHANICS.  239 

will   be   greater   than  that   upon  the   longer  column,  and 
this  excess  pressure  will  produce  motion. 

605.  Let  A  —  the  area  of  the  tube ; 

h  =  D  C  =  the  vertical  distance  between  the 
surface  of  the  water  in  />,  and  the  highest 
point  of  the  center  line  of  the  tube ; 

/*,  =  E  C  =  the  distance  between  the  surface 
of  the  water  in  A,  and  the  highest  point  of 
the  center  line  of  the  tube. 

The  weight  of  the  water  in  the  short  column  is  .03617  X 
A  h^  and  the  resultant  atmospheric  pressure  tending  to 
force  the  water  up  the  short  column  is  14.7  X  A  —  .03617  X 
A  hv.  The  weight  of  the  water  in  the  long  column  is 
.03617  A  //,  and  the  resultant  atmospheric  pressure  tending 
to  force  the  water  up  the  long  column  is  14. 7  A  —.03617^4  h. 
The  difference  between  these  two  is  (14.7  A  —  .03617  A  //,) 
—  (14.7  A  —  .03617  A  //)  =  .03617  A  (h  —  //,).  But  //  —  //,= 
E  D  —  the  difference  between  the  levels  of  the  water  in  the 
two  vessels.  In  the  above,  h  and  //,  were  taken  in  inches, 
and  A  in  square  inches. 

It  will  be  noticed  that  the  short  column  must  not  be 
higher  than  34  feet  for  water,  or  the  siphon  will  not  work, 
since  the  pressure  of  the  atmosphere  will  not  support  a 
column  of  water  that  is  higher  than  34  feet. 

606.  The  Injector. — A  section  of  an  injector  is  shown 
in  Fig.  127.     There  are  many  different  kinds  of  these  in- 
struments, but  the  principle  is  the  same  in  all.     They  are 
used  for   lifting   water  from  a  point  below  the   discharge 
orifice,  and  forcing  it  into  the  boiler  of  a  steam  engine  or 
locomotive.    They  depend  for  their  action  upon  the  creation 
of  a  partial  vacuum  by  the  action  of   steam.     The  valve 
A  is  opened  by  turning  the  hand  wheel  S,  and  the  steam 
enters  from  F  into  D,  and  flows  through  the  tubes  C,  K,  M, 
and  the  outlet  N,  into  the  boiler.     The  valve  B  is  opened 
by  turning  the  hand  wheel  W,     The  steam,  flowing  through 
C  with  a  high  velocity,  drives  out  the  air.     The  air  in  the 


240  MECHANICS. 

chamber  G  G  and  in  the  suction  pipe  P  expands  and  flows 
into  C  through  the  conical  orifice  //,  thus  creating  a  partial 
vacuum  in  the  chamber  G  and  pipe  P.  The  atmospheric 


pressure  on  the  water  causes  it  to  rise  in  pipe  Pt  and  flow 
into  G,  and  from  thence  into  C,  through  the  conical  orifice 
//,  whence  the  steam  drives  it  through  K,  Jlf,  and  N  into 
the  boiler. 

6O7.  The  method  of  operating  the  injector  is  as  follows: 
The  water  valve  B  is  opened;  next  the  small  valve  R  is 
opened  by  turning  the  handle  J.  The  steam  then  flows  into 
the  passage  £,  which  connects  with  the  chamber  D  through 
the  conical  tube  C,  and  creates  a  partial  vacuum  in  G,  as 
before  described.  The  quantity  of  steam  admitted  by  the 
valve  R  not  being  sufficient  to  force  all  of  the  water  which 
flows  through  the  orifice  H  into  the  boiler,  the  water  will 
accumulate  in  the  chamber  7",  raise  the  valve  Z,  and  flow 
down  through  u  and  out  at  the  overflow  outlet  O.  As  soon 
as  the  water  appears  at  O  the  valve  A  is  opened  as  far  as 
possible,  and  the  valve  R  closed.  If  the  water  still  flows 
out  at  O  close  the  valve  B  slightly  until  it  stops.  An  in- 
jector will  lift  water,  according  to  temperature  and  steam 
pressure,  from  6  to  26  feet. 


MECHANICS. 


241 


PUMPS. 

6O8.  The  Suction  Pump.  —  A  section  of  an  ordinary 
suction  pump  is  shown  in  Fig.  128.  Suppose  the  piston  to 
be  at  the  bottom  of  the  cylinder,  and  to  be  just  on  the  point 
of  moving  upwards  in  the  direction  of  the  arrow.  As  the 
piston  rises  it  leaves  a  vacuum  behind  it,  and  the  atmos- 
pheric pressure  upon  the  surface  of  the  water  in  the  well 
causes  it  to  rise  in  the  pipe  P  for  the  same  reason  that  the 
mercury  rises  in  the  barometer  tube.  The  water  rushes  up 
the  pipe  and  lifts  the  valve  V,  filling  the  empty  space  in  the 
cylinder  B  displaced  by  the  piston.  When  the  piston  has 
reached  the  end  of  its  stroke  the  water  entirely  fills  the 
space  between  the  bottom  of  the  piston  and  the  bottom  of 
the  cylinder,  and  also  the  pipe  P.  The  instant  that  the  pis- 
ton begins  its  down  stroke,  the  water  in  the  chamber  B 
tends  to  fall  back  into  the  well,  and  its  weight  forces  the 
valve  V  to  its  seat,  thus  preventing  any  downward  flow  of 
the  water.  As  the 
piston  descends  the 
water  must  give  way 
to  it,  and,  since  the 
valve  V  is  closed,  the 
valves  «,  u  must  open, 
and  thus  allow  the 
water  to  pass 
through  the  piston, 
as  shown  in  the  right- 
hand  figure.  When 
the  piston  has 
reached  the  end  of  its 
downward  stroke,  the 
weight  of  the  water  E 
above  closes  the 


valves  ;/, 


All  the 


water  resting  on  the  top  of  the  piston  is  then  lifted  with  the 
piston  on  its  upward  stroke,  and  discharged  through  the 
spout  Ay  the  valve  V  again  opening,  and  the  water  filling 
the  space  below  the  piston  as  before. 


242 


MECHANICS. 


6O9.  It  is  evident  that  the  distance  between  the  piston 
when  at  top  of  its  stroke  and  the  surface  of  the  water  in  the 
well  must  not  exceed  34  feet,  the  highest  column  of  water 
which  the  pressure  of  the  atmosphere  will  sustain,  since, 
otherwise,  the  water  in  the  pipe  would  not  rise  up  and  fill 
the  cylinder  as  the  piston  ascended.  In  practice,  this  dis- 
tance should  not  exceed  28  feet.  This  is  due  to  the  fact 
that  there  is  a  little  air  left  between  the  bottom  of  the 
piston  and  the  bottom  of  the  cylinder,  a  little  air  leaks 
through  the  valves,  which  are  not  perfectly  air-tight,  and  a 
pressure  is  needed  to  raise  the  valve  against  its  weight, 
which,  of  course,  acts  downwards.  There  are  many  varie- 
ties of  the  suction  pump,  differing  principally  in  the  valves 
and  piston,  but  the  principle  is  the  same  in  all. 

61 0.     The    Lifting    Pump.— 

A  section  of  a  lifting  pump  is  shown 
in  Fig.  129  These  pumps  are  used 
when  water  is  to  be  raised  to  greater 
heights  than  can  be  done  with  the 
ordinary  suction  pump.  As  will  be 
perceived,  it  is  essentially  the  same 
as  the  pump  previously  described, 
except  that  the  spout  is  fitted  with 
a  cock  and  has  a  pipe  attached  to  it, 
leading  to  the  point  of  discharge. 
If  it  is  desired  to  discharge  the  water 
at  the  spout,  the  cock  may  be  opened; 
otherwise,  the  cock  is  closed,  and 
the  water  is  lifted  by  the  piston  up 
through  the  pipe  P  to  the  point  of 
discharge,  the  valve  C  preventing  it 
from  falling  back  into  the  pump, 
FIG  129.  and  the  valve  V  preventing  the 

water  in  the  pump  from  falling  back  into  the  well. 

611.  Force  Pumps. — The  force  pump  differs  from 
the  lifting  pump  in  several  important .  particulars,  but 
chiefly  in  the  fact  that  the  piston  is  solid  ;  that  is,  it  has 


MECHANICS. 


243 


no  valves.  A  section  of  a  suction  and  force  pump  is  shown 
in  Fig.  130.  The 
water  is  drawn  up 
the  suction  pipe  as 
before,  when  the 
piston  rises;  but 
when  the  piston  re- 
verses, the  pressure 
on  the  water  caused 
by  the  descent  of  the 
piston  opens  the 
valve  V  and  forces 
the  water  up  the  de- 
livery pipe  P' .  When 
the  piston  again  be- 
gins  its  upward  FIG.  130. 

movement,  the  valve  V  is  closed  by  the  pressure  of  the 
water  above  it,  and  the  valve  V  is  opened  by  the  pres- 
sure of  the  atmosphere  on  the  water  below  it,  as  in  the 
previous  cases.  For  an  arrangement  of  this  kind,  it  is 
not  necessary  to  have  a  stuffing-box.  The  water  may  be 
forced  to  almost  any  desired  height.  The  force  pump  differs 
again  from  the  lifting  pump  in  respect  to  its  piston  rod, 
which  should  not  be  longer  than  is  absolutely  necessary  in 
order  to  prevent  it  from  buckling,  while  in  the  lifting  pump 
the  length  of  the  piston  rod  is  a  matter  of  indifference. 

612.  Plunger  Pumps. — When  force  pumps  are  used 
to  convey  water  to  great  heights,  the  pressure  of  the  water 
in  the  cylinder  becomes  so  great  that  it  becomes  extremely 
difficult  to  keep  the  water  from  leaking  past  the  piston,  and 
the  constant  repairing  of  the  piston  packing  becomes  a 
nuisance.  To  obviate  this  difficulty  the  piston  is  made  very 
long,  as  shown  in  Fig.  131,  and  is  then  called  the  plunger. 
The  suction  valve  in  this  case  consists  of  two  clack  valves 
inclined  to  each  other  and  resting  upon  a  square  pin  A ; 
they  are  prevented  from  flying  back  too  far  during  the  up 
stroke  of  the  plunger  by  the  two  uprights  /,  /.  During  the 


244 


MECHANICS. 


down  stroke  of  the  plunger,  the  valves  at  A  are  closed  and 
the  valve  B  in  the  delivery  pipe  is  open  A  little  air  is  always 
carried  into  the  cylinder  of  a 
pump  with  the  entering  water. 
In  force  pumps  this  fact  becomes 
a  serious  consideration,  since 
after  repeated  strokes  the  air 
accumulates,  and  during  the  down 
stroke  of  the  plunger  it  is  com- 
pressed. After  a  time  it  would 
become  sufficient  to  entirely  pre- 
vent the  water  from  entering 
through  the  suction  valve,  the 
pressure  on  the  top  of  the  valve 
being  greater  than  that  of  the 
atmosphere  below.  In  the  pump 
shown  in  the  figure,  the  plunger 
is  a  trifle  smaller  than  the  cylin- 
der, and  the  air  collects  around 
the  plunger  below  the  stuffing- 
box.  To  remove  this  air  a  narrow 
passage  C  (shown  by  the  dotted 
FIG-  131-  lines),  that  can  be  closed  at  its 

upper  end  by  the  cock  Z>,  connects  the  interior  of  the 
pump  with  the  atmosphere  when  the  cock  is  open.  It 
is  evident  that  this  cock  must  not  be  opened  except  dur- 
ing the  down  stroke  of  the  plunger,  for,  if  it  were  open 
during  the  up  stroke,  the  pressure  below  the  plunger  being 
less  than  the  pressure  of  the  atmosphere  above>  the  air  would 
rush  in  instead  of  being  expelled. 


AIR    CHAMBERS. 

613.  In  order  to  obtain  a  continuous  flow  of  water  in 
the  delivery  pipe,  with  as  nearly  a  uniform  velocity  as  pos- 
sible, an  air  chamber  is  usually  placed  on  the  delivery 
pipe  of  force  pumps  as  near  to  the  pump  cylinder  as  the  con- 
struction of  the  machine  will  allow.  The  air  chambers  are 
usually  pear-shaped,  with  the  small  end  connected  to  the 


MECHANICS. 


245 


pipe.  They  are  filled  with  air,  which  the  water  compresses 
during  the  discharge.  During  the  suction,  the  air  thus 
compressed  expands  and  acts  as  an  accelerating  force  upon 
the  moving  column  of  water,  a  force  which  diminishes  with 
the  expansion  of  the  air  and  helps  to  keep  the  velocity  of  the 
moving  column  more  nearly  uniform.  An  air  chamber 
is  sometimes  placed  upon  the  suction  pipe.  These  air  cham- 
bers not  only  tend  to  promote  a  uniform  discharge,  but  also 
to  equalize  the  stresses  upon  the  pump  and  prevent  shocks 
due  to  the  incompressibility  of  water.  They  subserve  the 
same  purpose  on  pumps  that  a  fly-wheel  does  on  the 
steam  engine.  Unless  the  pump  moves  very  slowly,  it  is 
absolutely  necessary  to  have  an  air  chamber  on  the  delivery 
pipe. 

STEAM     PUMPS. 

614.  Steam  pumps  are  force  pumps  operated  by  steam 
acting  upon  the  piston  of  a  steam  engine  directly  connected 
to  the  pump,  and  in  many  cases  cast  with  the  pump.  A 


section  of  a  double-acting  steam  pump,  showing  the  steam 
cylinders,  with  other  details,  is  illustrated  in 
HWp  CT  i«s  the  steam  niston.  and  R  the  niston  rod. 


and 
Fig. 


water 
132.     Here 


lers,    wim   otner  aetans,  is  mustraiea  in 
rj  is  the  steam  piston,  and  R  the  piston  rod, 


246  MECHANICS. 

which  is  secured  at  its  other  end  to  the  pump  plunger  P. 
F  is  a  partition  cast  with  the  cylinder,  which  prevents  the 
water  in  the  left-hand  half  from  communicating  with  that 
in  the  right-hand  half  of  the  cylinder.  Suppose  the  piston 
to  be  moving  in  the  direction  of  the  arrow.  When  the  pis- 
ton has  arrived  at  the  end  of  its  stroke,  the  water  space  in 
the  left-hand  half  of  the  pump  cylinder  will  have  been  in- 
creased by  an  amount  equal  to  the  area  of  the  cross-section 
of  the  plunger,  multiplied  by  the  length  of  the  stroke,  and 
the  volume  of  the  right-hand  half  of  the  cylinder  will  have 
been  diminished  by  a  like  amount.  In  consequence  of  this, 
a  volume  of  water  in  the  right-hand  half  of  the  cylinder 
equal  to  the  volume  displaced  by  the  plunger  in  its  forward 
movement  will  be  forced  through  the  valves  f7,'  V,  through 
the  orifice  D,  into  the  air  chamber  A,  and  then  discharged 
through  the  delivery  pipe  H.  By  reason  of  the  partial 
vacuum  in  the  left-hand  half  of  the  pump  cylinder,  owing  to 
this  movement  of  the  plunger,  the  water  will  be  drawn  from 
the  reservoir  through  the  suction  pipe  C  into  the  chamber 
K  K,  lifting  the  valves  5',  S'  and  filling  the  space  displaced 
by  the  plunger.  During  the  return  stroke  the  water  will  be 
drawn  through  the  valves  S,  S  into  the  right-hand  half  of 
the  pump  cylinder,  and  at  the  same  time  water  will  be  dis- 
charged from  the  left-hand  half  through  the  valves  F,  F/out 
through  the  pipe  //,  as  before.  Each  one  of  the  four  suc- 
tion and  four  discharge  valves  is  kept  to  its  seat,  when  not 
working,  by  light  springs,  as  shown, 

615.  There  are  many  varieties  and  makes  of  steam 
pumps,  the  majority  of  which  are  double-acting.  In  many 
cases  two  steam  pumps  are  placed  side  by  side,  having  a 
common  delivery  pipe.  This  arrangement  is  called  a 
duplex  pump.  It  is  usual  to  so  set  the  steam  pistons  of 
duplex  pumps  that  when  one  is  completing  the  stroke,  the 
other  is  in  the  middle  of  its  stroke.  A  double-acting 
duplex  pump  made  to  run  in  this  manner,  and  having  an 
air  chamber  of  sufficient  size,  will  deliver  water  with  nearly 
a  uniform  velocity. 


MECHANICS.  247 

In  mine  pumps  for  forcing  water  to  great  heights,  the 
plungers  are  made  solid,  and  in  most  cases  extend  through 
the  pump  cylinder.  In  many  steam  pumps,  pistons  are 
used  instead  of  plungers,  but  when  very  heavy  duty  is 
required,  plungers  are  preferred. 


STRENGTH  OF   MATERIALS. 

616.  When    a   force  is  applied  to  a  body,  it  changes 
either  its  form  or  its  volume.     A  force,  when   considered 
with  reference  to  the  internal  changes  it  tends  to  produce 
in  any  solid,  is  called  a  stress. 

Thus,  if  we  suspend  a  weight  of  2  tons  by  a  rod,  the  stress 
in  the  rod  is  2  tons.  This  stress  is  accompanied  by  a 
lengthening  of  the  rod,  which  increases  until  the  internal 
stress  or  resistance  is  in  equilibrium  with  the  external 
weight. 

617.  Stresses  may  be  classified  as  follows: 


Tensile,  or  pulling  stress. 
Compressive,  or  pushing 


stress. 


Transverse,  or  bending  stress. 
Shearing,  or  cutting  stress. 
Torsional,  or  twisting  stress. 

618.  A  unit  stress  is  the  amount  of  stress  on  a  unit 
of  area,  and  may  be  expressed  either  in  pounds  per  square 
inch,  or  in  tons  per  square  foot;  or,  it  is  the  load  per  square 
inch  or  square  foot  on  any  body. 

Thus,  if  10  tons  are  suspended  by  a  wrought-iron  bar 
which  has  an  area  of  5  square  inches,  the  unit  stress  is 

2  tons  per  square  inch,  because  —  =  2  tons. 

o 

619.  Strain  is  the  deformation  or  change  of  shape  of 
a  body  resulting  from  stress. 

For  example,  if  a  rod  100  feet  long  is  pulled  in  the  direc- 
tion of  its  length,  and  if  it  is  lengthened  1  foot,  it  is 
strained  yf^th  of  its  length,  or  1  per  cent. 


62O.     Elasticity  is  the  power  which  a  body  has  of 
returning  to  its  original  form  after  the  external  force  on 


248  MECHANICS. 

it  is  withdrawn,  providing  the  stress  has  not  exceeded  the 
elastic  limit. 

Consequently,  we  see  from  this  that  all  material  is 
lengthened  or  shortened  when  subjected  to  either  tensile 
or  compressive  stress,  and  the  change  of  the  length  is 
directly  proportional  to  the  stress,  within  the  elastic  limit. 

For  stresses  within  the  elastic  limits,  materials  are  per- 
fectly elastic,  and  return  to  their  original  length  on  removal 
of  the  stresses;  but,  when  their  elastic  limits  are  exceeded, 
the  changes  of  their  lengths  are  no  longer  regular,  and  a 
permanent  set  takes  place;  the  destruction  of  the  material 
has  then  begun. 

621.  The  measure  of  elasticity  of  any  material  is 
the  change  of  length  under  stress  within  the  elastic  limit. 

622.  The    elastic    limit    is   that   unit    stress    under 
which  the  permanent  set  becomes  visible. 

The  elasticity  of  wrought  iron  is  practically  the  same  as 
that  of  steel;  that  is,  each  material  will  change  an  equal 
amount  of  length  under  the  same  stress  within  the  elastic 
limits. 

The  elastic  limit  of  steel  is  higher  than  that  of  wrought 
iron;  consequently,  the  former  will  lengthen  or  shorten 
more  than  the  latter  before  its  elasticity  is  injured. 


TENSILE    STRENGTH    OF    MATERIALS. 

623.  The  tensile  strength  of  any  material  is  the 
resistance  offered  by  its  fibers  to  being  pulled  apart. 

The  tensile  strength  of  any  material  is  proportional  to 
the  area  of  its  cross-section. 

Consequently,  when  it  is  required  to  find  the  safe  tensile 
strength  of  any  material,  we  have  only  to  find  the  area  at 
the  minimum  cross-section  of  the  body,  and  multiply  it  by 
its  strength  per  square  inch,  as  given  in  the  following  table 
under  the  heading  "Working  Stress." 

NOTE. — The  minimum  cross-section  referred  to  in  the  above  para- 
graph is  that  section  of  the  material  which  is  pierced  with  holes;  such 
as  bolt  or  rivet  holes  in  iron,  or  knots  in  wood,  if  there  are  any. 


MECHANICS. 


249 


624.     In  the  following  table  is  given  the  average  break- 
ing and  working  tensile  stress  of  different  materials: 


TABLE   17. 


Material. 

Breaking  Stress 
in  Pounds  per  Square 
Inch. 

Working  Stress 
in  Pounds  per  Square 
Inch. 

Timber  
Cast  Iron  
Wrought  Iron  
Steel  

10,000 
16,000 
50,000 
70  000 

600  to     1,200 
1,500  to    3,500 
5,000  to  12,000 
6  000  to  13  000 

The  above  table  shows  that  the  tensile  breaking  strength 
of  cast  iron  is  16,000  pounds  per  square  inch  of  cross-section, 
and  that  the  working  strength  is  from  1,500  to  3,500  pounds 
per  square  inch  of  cross-section. 

625.  In  machinery,  such  as  steam  engines,  where  the 
parts  are  subjected  to  shocks,  or  are  alternately  compressed 
and  extended,  it  is  not  safe  to  subject  cast  iron  to  a  stress 
of  more  than  1,500  pounds  per  square  inch  of  section, 
wrought  iron  to  more  than  5,000  pounds  per  square  inch  of 
section,  or  steel  to  more  than  6,000  pounds  per  square  inch 
of  section. 

But  in  structures  in  which  the  strains  are  constantly  in 
one  direction,  as  is  the  case  with  steam  boilers,  wrought 
iron  may  be  strained  with  from  6,000  to  8,000  pounds  per 
square  inch  of  section,  or  steel  with  from  8,000  to  10,000 
pounds  per  square  inch  of  section. 

Consequently,  strict  attention  must  be  given  to  the  nature 
of  the  load  the  given  structure  has  to  bear,  and  fix  the 
working  stress  accordingly. 

NOTE. — For  structures  on  which  the  load  is  applied  suddenly,  use 
the  smaller  working  stresses  given  in  the  table,  ana  for  those  on  which 
the  load  is  applied  gradually,  use  the  larger  working  stresses. 


250  MECHANICS. 

RULES    AND    FORMULAS    FOR    TENSILE 

STRENGTH. 
626.     Let  W  —  safe  load  in  pounds; 

A  —  area  of  minimum  cross  -section; 
5"  =  working    stress   in   pounds  per   square 
inch,  as  given  in  the  foregoing  table. 

Rule  111.  —  The  working  load  in  pounds  for  any  bar 
subjected  to  a  tensile  stress  is  equal  to  the  minimum  sectional 
area  of  the  bar,  multiplied  by  the  working  stress  in  pounds 
per  square  inch,  as  given  in  the  table. 

That  is,  W=A  S. 

EXAMPLE.  —  A  bar  of  good  wrought  iron  which  is  3"  square  is  to  be 
subjected  to  a  steady  tensile  stress  ;  what  is  the  maximum  load  that  it 
should  carry? 

SOLUTION.  —  From  what  has  been  said  above  in  regard  to  the 
materials  and  to  the  nature  of  the  load,  it  will  be  safe  in  this  case  to 
use  a  working  stress  of  12.000  pounds  per  square  inch. 

Applying  the  rule,  we  have 

=  108,000  pounds.     Ans. 


Rule  112.  —  The  •minimum  sectional  area  of  any  bar 
subjected  to  a  tensile  stress  should  be  equal  to  the  load  in 
pounds,  divided  by  the  working  stress  in  pounds  per  square 
inch,  as  given  in  the  table. 

W 

That  is,  A  =  -^. 
o 

EXAMPLE.  —  What  should  be  the  area  of  a  wrought-iron  bar  to  carry 
a  steady  load  of  108,000  pounds,  if  it  is  to  resist  a"  tensile  stress  of 
12,000  pounds  per  square  inch  ? 

SOLUTION.  —  Applying  the  rule, 


Rule  1  1  3.  —  The  working  stress  in  pounds  per  square 
inch  is  equal  to  the  load  in  pounds  divided  by  the  minimum 
sectional  area  of  the  bar. 

W 
That  is,  S  =       . 


MECHANICS.  251 

EXAMPLE.— A  bar  of  wrought  iron  3"  square,  subjected  to  tensile 
stress,  carries  a  load  of  108,000  pounds;  what  is  the  stress  per  square 


inch  : 


SOLUTION.— Applying  the  rule, 


S  =  1f'(Xf)  =  12,000  Ib.  per  sq.  in.     Ans. 
o  X  o 


CHAINS. 

627.  Chains  made  of  the  same  size  iron  vary  in  strength, 
owing  to  the  different  kinds  of  links  from  which  they  are 
made. 

It  is  a  good  practice  to  anneal  old  chains  which  have 
become  brittle  by  overstraining.  This  renders  them  less 
liable  to  snap  from  sudden  jerks.  It  reduces  their  tensile 
strength,  but  increases  their  toughness  and  ductility,  which 
are  sometimes  more  important  qualities. 

When  annealing,  care  should  be  taken  that  a  sufficient 
heat  be  applied,  otherwise  no  benefit  will  be  gained;  the 
chains  ought  to  be  heated  to  a  cherry  red,  say  1300°  F.  at 
the  least. 

628.  Rules   and   Formulas  for  the   Strength  of 
Chains : 

Let  W '=  safe  load  in  pounds; 

D  =  diameter  of  the  iron,  in  inches,  from  which  the 
links  are  made. 

Rule  114. —  The  safe  load  in  pounds  of  a  stud-link 
wrought-iron  chain  is  equal  to  18,000,  multiplied  by  the 
square  of  the  diameter  of  the  iron  from  which  the  links  are 
made. 

That  is,  W=  18,000  D\ 

EXAMPLE.— What  is  the  maximum  load  that  should  be  carried  by  a 
stud-link  wrought-iron  chain,  if  its  links  are  made  from  J-inch  round 


iron 


SOLUTION.— Applying  the  rule,   W  =  18,000  D*. 
Substituting  the  value  of  Z?«,  we  have  W=  18,000  x  J  X  |-  =  10,125 
pounds.    Ans. 


252  MECHANICS. 

Rule  115. —  The  safe  load  in  pounds  of  a  close-link 
wrougJit-iron  chain  is  equal  to  13,000,  multiplied  by  the 
square  of  the  diameter  of  the  iron  from  which  the  links  are 
made. 

That  is,  W=  12,000  D\ 

EXAMPLE.— What  is  the  maximum  load  that  should  be  carried  by  a 
close-link  wrought-iron  chain,  if  its  links  are  made  from  f-inch  round 
iron? 

SOLUTION.— Applying  the  rule,   W '=  IS.OOOZ?2. 

O  O 

Substituting  the  value  of  Z>*,  we  have  W=  12,000  X  -j  x  T  = 
pounds.     Ans. 


HEMP   ROPES. 

629.  The  strength  of  hemp  ropes  does  not  depend  so 
much  upon  the  quality  of  the  material  and  the  cross-section 
of  the  rope,  as  upon  the  method  of  manufacture  and  the 
amount  of  twisting. 

The  ropes  in  common  use  are  three-strand,  shroud-laid 
rope,  and  hawser  or  cable-laid  rope. 

The  strongest  ropes  are  three-strand  shroud-laid,  made 
without  tar.  Ropes  made  with  tar  are  less  flexible,  and  are 
reduced  in  strength  about  25  per  cent.,  but  have  better 
wearing  qualities. 

630.  Rules  and  Formulas  for   the   Strength   of 
Hemp  Ropes  : 

Let  W=  maximum  working  load  in  pounds; 
C  =  circumference  of  rope  in  inches. 

Rule  116. —  The  maximum  -working  load  in  pounds  that 
should  be  allowed  on  any  liemp  rope  is  equal  to  the  square  of 
the  circumference  of  the  rope,  multiplied  by  100. 

That  is,  IV  =100  C\ 

EXAMPLE. — What  is  the  maximum  load  in  pounds  that  should  be 
carried  by  a  hemp  rope  which  has  a  circumference  of  8  inches  ? 

SOLUTION.— Substituting  the  value  of  Cin  the  formula,  W=  100  x 
8*  =  6,400  Ib.  Ans. 


MECHANICS.  253 

Rule  117. —  The  circumference  of  any  Jiemp  rope  is  equal 
to  the  square  root  of  the  maximum  working  load  in  pounds 
which  it  is  capable  of  carrying,  multiplied  by  .  1. 
That  is,  C=.l  /IF. 

EXAMPLE.— A  maximum  working  load  of  1,000  pounds  is  to  be  carried 
by  a  hemp  rope;  what  should  be  the  circumference  of  the  rope  ? 

SOLUTION.— Applying  the  rule,  C  =  .1  yXOOO  =  3.16".     Ans. 

When  measuring  ropes,  the  circumference  is  sought 
instead  of  the  diameter,  because  the  ropes  are  not  round 
and  the  circumference  therefore  is  not  3.1416  times  the 
diameter.  For  three  strands  the  circumference  is  about 
2.86  d\  for  seven  strands,  3  d. 

631.  The  above  formulas  are  very  convenient  for  use, 
and  easily  remembered,  but  it  is  well  to  remark  that  the 
values  thus  given  apply  to  ropes  of  a  good  average  quality. 

WIRE  ROPES. 

632.  Wire  rope  is  made  of  iron  and  steel  wire.     It  is 
stronger  than  hemp  rope,  and,  to  carry  the  same  load,  is  of 
smaller  diameter. 

In  substituting  steel  for  iron  rope,  the  object  in  view 
should  be  to  gain  an  increase  of  wear  from  the  rope,  rather 
than  to  reduce  the  size. 

A  steel  rope  to  be  serviceable  should  be  of  the  best  obtain- 
able quality,  because  ropes  made  from  low  grades  of  steel 
are  inferior  to  good  iron  ropes. 

633.  Formulas  for  the  Strength  of  Wire  Ropes  : 

Let  W-=  maximum  of  working  load  in  pounds; 

C  =  circumference  of  rope  in  inches. 

Rule   118. —  The  maximum  working  load  in  pounds  that 
should  be  allowed  on  any  iron  wire  rope  is  equal  to  the  square 
of  the  circumference  of  the  rope  in  inches,  multiplied  by  600. 
That  is,  W=  600  C2. 

EXAMPLE. — What  is  the  maximum  load  in  pounds  that  should  be 
carried  by  an  iron  wire  rope  whose  circumference  is  4.}  inches  ? 
SOLUTION.— Applying  the  formula. 

^=600X4.5^  =  12,1501^     Ans. 


254  MECHANICS. 

Rule  119. —  The  circumference  of  any  iron  wire  rope  in 
inches  is  equal  to  the  square  root  of  the  maximum  working 
load  in  pounds,  multiplied  by  .0408. 

That  is,  <r  =  .0408VTF. 

EXAMPLE.— A  maximum  working  load  of  12,150  pounds  is  to  be  car- 
ried by  an  iron  wire  rope;  what  should  be  the  minimum  circumference 
of  the  rope  ? 

SOLUTION.— Applying  the  formula, 

C  =  .0408  4/12,150  =  4^  inches.     Ans. 

Rule  12O. —  The  above  rules  and  formulas  are  also  made 
applicable  when  computing  the  safe  strength  of  steel  wire 
rope,  by  substituting  the  constant  1,000  for  the  constant  600, 
and  .0316  for  .0408. 

EXAMPLE. — What  is  the  maximum  load  in  pounds  that  should  be 
carried  by  a  steel  wire  rope,  the  circumference  of  which  is  4^  inches  ? 
SOLUTION.— Applying  the  rule,   W  —  1,000  X  4.5'  =  20,250  Ib.     Ans. 

EXAMPLE. — A  maximum  wor-king  load  of  10,485  pounds  is  to  be  car- 
ried by  a  steel  wire  rope;  what  should  be  the  minimum  circumference 
of  the  rope  ? 

SOLUTION.— Applying  the  rule,  (7  =  .0316  4/10,485  =  3.24  inches. 

Ans. 


EXAMPLES   FOR   PRACTICE. 

1.  What  should  be  the  diameter  of  a  steel  piston  rod  of  a  steam  en- 
gine to  resist  tension,  if  the  piston  is  19"  in  diameter  and  the  pressure 
is  85  Ib.  per  sq.  in.  ?  Ans.  2^",  nearly. 

2.  What  safe  load  will  a  cast-iron  bar  of  rectangular  cross-section 
1$"  by  3i"  support  if  subjected  to  shocks  ?    The  bar  is  in  tension. 

Ans.  39,375  Ib. 

3.  What  is  the  stress  per  sq.  in.  on  a  piece  of  timber  8"  square, 
which  is  subjected  to  a  steady  pull  of  60,000  pounds  ? 

Ans.  937.5  Ib.  per  sq.  in. 

4.  What  should  be  the  safe  load  for  a  close-link  wrought-iron  chain 
whose  links  are  made  from  $"  iron  ?  Ans.  9,187.5  Ib. 

5.  What  safe  load  may  a  hemp  rope  carry  whose  circumference  is 
4"?  Ans.  1,600  Ib. 

6.  What  should  be  the  allowable  working  load  for  a  steel  wire  rope 
whose  circumference  is  3f  ?  Ans.   14,062.5  Ib. 

7.  What  should  be  the  circumference  of  an  iron  wire  rope  to  sup- 
port a  load  of  20,000  Ib.  ?  Ans.  5|",  nearly. 


MECHANICS. 


255 


CRUSHING    STRENGTH    OF    MATERIALS. 

634.  The  crushing  strength  of  any  material  is  the 
resistance  offered  by  its  fibers  to  being  pushed  together. 

To  obtain  only  compression,  the  length  of  a  rod  should 
not  be  more  than  5  times  greater  than  its  least  diameter, 
or  its  least  thickness  when  it  is  a  rectangular  rod. 

If  a  bar  is  long  compared  with  its  cross  dimensions,  the 
load  if  sufficiently  great  will  cause  it  to  bend  sideways 
under  the  compressive  force,  and  we  have,  then,  not  only 
compression,  but  compression  compounded  with  bending. 

Experimental  tests  on  pillars  have  shown  that  their 
strengths  are  approximately  inversely  proportional  to  the 
squares  of  their  lengths.  That  is,  if  there  are  two  pillars 
of  the  same  material,  having  the  same  cross-section,  but 
one  is  twice  as  long  as  the  other,  the  long  one  will  sustain 
only  about  one-quarter  the  load  of  the  short  one. 


635.     Attention  should  be  given  to  the  ends  of  pillars, 
as   their   shape  has  great   influ-  w^m 
ence    upon   their   strength.      In  y, 
Fig.  133  are  shown  three  pillars  "^ 
with  differently  shaped  ends 

It  has  been  proven  by  the  aid 
of  higher  mathematics  that, 
theoretically,  a  pillar  having  flat 
or  fixed  ends,  as  shown  at  a,  is 
four  times  as  strong  as  one  that 
has  round  or  movable  ends,  as 
shown  at  c,  and  one  and  seven-  4. 
ninths  times  as  strong  as  one  i 
having  one  flat  and  one  round 

end,  as  shown  at  b\  b  is  thus  two  and  one-fourth  times  as 
strong  as  c.  It  has  also  been  found  that  if  three  pillars, 
a,  b^  c,  which  have  the  same  crossrsection,  are  to  carry  the 
same  load  and  be  of  equal  strength,  their  lengths  must  be 
as  the  numbers  2,  \\,  and  1,  respectively. 

In  practice,  however,  the  ends  of  the  pillars  b  and  rare  not 
generally  made  as  shown  by  the  figure,  but  have  holes  at 


256 


MECHANICS. 


their  ends  into  which  pins  are  fitted  which  are  fastened  to 
some  other  piece ;  as,  for  example,  the  connecting-rod  of  an 
engine.  In  such  cases,  it  has  been  found  that  a  is  two 
times  as  strong  as  c,  and  that  b  is  one  and  one-half  times  as 
strong  as  c.  That  is,  in  actual  practice,  a  column  fixed  as 
at  c  is  really  |-  as  strong  as  one  fixed  as  at  a,  instead  of 
being  only  £  as  strong,  as  given  above. 

Green  or  wet  timber  has  only  one-half  the  strength  of  dry 
and  seasoned  timber;  consequently,  its  crushing  strength  is 
only  one-half  of  that  given  in  the  table  below. 

636.  In  the  following  table  is  given  the  mean  crushing 
strength  of  some  short  specimens  of  materials  in  tons  (of 
2,000  pounds)  per  square  inch: 

TABLE   18. 


Materials. 

Crushing 
Strength  in 
Tons  per 
Square  Inch. 

Cast  Iron 

40 

Wrought  Iron  

18 

Mild  Steel 

26 

Cast  Copper.  .  .  . 

5 

Cast  Brass  
Timber  (dry)  endwise  
Brick  

4.5 
3.5 
1 

Stone 

3 

637.     Formula  for  the  Strength  of  Pillars: 

The  following  formula  is  applicable  to  pillars  which  are 
commonly  used  in  practice,  the  lengths  of  which  are  about 
from  10  to  40  times  their  least  diameter,  or,  if  rectangular, 
their  least  thickness  as  indicated  by  d. 

Let  C=  crushing  strength  in  tons  per  square  inch  of  a 
short  specimen  of  the  material,  as  given  in  the 
above  table; 

5  =  sectional  area  in  square  inches; 
L  =  length  in  inches; 


MECHANICS. 


257 


d  =  least  thickness  of  rectangular  pillar,  or  diameter 

of  round  pillar  in  inches; 
W=  breaking  load  in  tons; 
A  =  the  area  of  the  two  flanges; 
B  =  the  area  of  the  web ; 
a  =  constant,    given  in  one   of   the  following  three 

tables: 

TABLE    19. 

CONSTANTS  FOR  WROUGHT-IRON  PILLARS. 


Cross-section  of  Pillar. 

When  Both 
Ends  of  the 
Pillar  are  Flat 
or  Fixed. 

When  One  End 
of  the  Pillar 
is  Flat  or  Fixed, 
and  the  Other 
Round  or 
Movable. 

When  Both 
Ends  of  the 
Pillar  are 
Round  or 
Movable. 

r*-i 

)       Round. 

2,250 

1,500 

1,125 

tdJ    h"< 

*-. 

3,000 

2,000 

1,500 

mm 

1   Square  or 
H   Rectangle. 

ii 

Thin  Square 
Tube. 

6,000 

4,000 

3,000 

Thin  Round 
Tube. 

4,500 

3,000 

2,250 

l_ 

Angle  with 
Equal  Sides. 

1,500 

1,000 

750 

l"~B~^      Cross  with 
bjpi      Equal  Arras 

1,500 

1,000 

750 

I 

I  Beam. 

A 

A 

—A 

A+B 

A+B 

258 


MECHANICS, 


TABLE   2O. 

CONSTANTS    FOR    CAST-IRON    PILLARS. 


Cross-section  of  Pillar. 

When  Both 
Ends  of  the 
Pillar  are  Flat 
or  Fixed. 

When  One  End 
of  the  Pillar  is 
Flat  or  Fixed, 
and  the  Other 
Round  or 
Movable. 

When  Both 
Ends  of  the 
Pillar  are 
Round 
or  Movable. 

Round. 

281.25 
375. 

187.5 
250. 

140.625 
187.5 

Ill             Square  or 
|    Rectangle. 

|P™l4     Thin  Square 

750. 

500. 

375. 

OThin  Round 
Tube. 

562  .  5 

375. 

281.25 

k^J     Angle  with 
Equal  Sides. 

187.5 

125. 

93.75 

\~-  d-^      Cross  with 
LJM      Equal  Arms. 

187.5 

125. 

93.75 

^"       I  Beam. 

A 

0?CA    v           _ 

125  X  -^g 

XX+B 

A  -\-B 

MECHANICS. 


259 


TABLE  21. 

CONSTANTS   FOR   WOODEN   PILLARS. 


Cross-section  of  Pillar. 

When  Both 
Ends  of  the 
Pillar  are   Flat 
or  Fixed. 

When  One  End 
of  the  Pillar  is 
Flat  or  Fixed, 
and  the   Other 
Round  or 
Movable. 

When  Both 
Ends  of  the 
Pillar  are 
Round 
or  Movable. 

Round. 

187.5 

125. 

93.75 

t-d]   f—  «H    Square 

250. 

166.66 

125. 

|j 

Hollow 

h 

I       Square 
Made  of 

500. 

333.33 

250. 

U 

I       Boards. 

638.  Rule  121.  —  The  breaking  load  of  a  pillar  in  tons  is 
equal  to  the  crushing  strength  of  a  short  specimen  of  the 
'material  as  given  in  Table  18,  Art.  636,  multiplied  by  the  sec- 
tional area  of  the  pillar  in  square  inches,  and  the  product 
divided  by  1  plus  the  quotient  obtained  by  dividing  the  square 
of  the  length  of  the  pillar  in  inches  by  the  square  of  the 
diameter  (or  least  thickness,  if  rectangular}  multiplied  by 

the  value  of  a. 

C  S 
That  is,  W=—  '-^ 


The  result  obtained  by  the  formula  must  be  divided  by 
6  to  get  the  safe  working  load. 

NOTE.—  If  the  length  of  the  pillar  is  given  in  feet,  be  sure  to  reduce 
it  to  inches  before  substituting  in  the  formula. 

EXAMPLE.  —  A  wooden  pillar,  6  inches  square  and  144  inches  long,  is 
fixed  at  both  ends;  what  load  will  it  sustain  with  safety  ? 

SOLUTION.—  By  rule  121,  W=  —  —  n. 


Substituting  the  values  of  C,  S,  Z*,  a,  and  d*,  we  have 


260  MECHANICS. 

3.5  X  6  X  6          00  . 

144  x  144   =  38-14  tons-  nearly. 

1  +  250  x  6  X  6 

Which,  divided  by  6,  gives  —  ^  —  =  6.357  tons,  or  the  load  it  is  capable 
of  sustaining  with  safety.  Ans. 

EXAMPLE.  —  A  wrought-iron  pillar,  4  inches  in  diameter  and  60  inches 
long,  is  fixed  at  one  end  and  movable  at  the  other  ;  what  load  will  it 
sustain  with  safety  ? 

SOLUTION.—  By  the  rule, 

W=  18X4X4*7854  =  196.69  tons. 


1,500x4x4 

Which,  divided  by  6,  gives  —  —  =  32.78  tons,  nearly,  or  the  load 
it  is  capable  of  sustaining  with  safety.  Ans. 

EXAMPLE.—  What  load  will  a  cast-iron  pillar  sustain  with  safety,  if 
it  is  20  feet  long,  and  if  its  cross-section  is  a  cross  with  equal  arms, 
two  of  which  are  equal  to  10  inches  in  length  (as  d,  see  table),  and 
whose  arms  are  1  inch  thick  ;  both  ends  of  the  pillar  movable. 

SOLUTION.  —  Area  of  cross-section  =  (10  x  1)  +  2(4.5  X  1)  =  19  square 
inches;  20  feet  =  240  inches. 


93.75x10x10 

Which,  divided  by  6,  gives  ^^  =  17.73  tons,  the  load  it  is  capable 
of  sustaining  with  safety.  Ans. 

639.  When  using  this  formula,  first  obtain  the  value  of 
Cirom  Table  18,  Art.  636.  Next,  calculate  the  area  of 
the  cross-section  of  the  pillar.  Then,  find  the  value  of  a 
from  one  of  the  last  four  tables.  Finally,  be  sure  that  the 
length  of  the  pillar  has  been  reduced  to  inches  before 
substituting  in  the  fprmula. 

To  find  the  proper  value  of  a  in  any  example,  first  turn  to 
the  table  dealing  with  the  material  in  question,  and  find  the 
figure  corresponding  to  the  given  cross-section  ;  in  the  hori- 
zontal line  containing  this  are  three  numbers  corresponding 
to  the  different  conditions  of  the  ends  of  the  column. 
From  these  numbers  select  the  one  corresponding  to  the 
given  conditions  of  the  column  to  be  calculated,  and  this 
will  be  the  required  value  of  a. 


MECHANICS. 


261 


EXAMPLES   FOR   PRACTICE. 

1.  What  load  may  be  safely  carried  by  a  hollow  cylindrical  cast- 
iron  pillar,  20  ft.  long,  inside  diameter  8".  and  outside  diameter   10"  ? 
Both  ends  of  the  pillar  are  fixed.  Ans.  93.13  tons. 

2.  A  rectangular  wooden  column  is  14  ft.  long,  and  has  one  end 
rounded;  if  the  cross-section  is  12"  x  8',  what  load  will  be  required  to 
break  it  ?  Ans.  92. 15  tons. 

3.  A  solid  wrought-iron  column,  which  has  both  ends  movable,  is  3" 
in  diameter  and  8  ft.  long;  what  load  will  it  safely  support  ? 

Ans.  11.1  tons. 


TRANSVERSE    STRENGTH    OF    MATERIALS. 

64O.  The  transverse  strength  of  any  material  is  the 
resistance  offered  by  its  fibers  to  being  broken  by  bending. 
As,  for  example,  when  a  beam,  bar,  rod,  etc.,  which  is  sup- 
ported at  its  ends,  is  broken  by  a  force  applied  between  its 
supports. 

The  transverse  strength  of  any  beam,  bar,  rod,  etc.,  is 
proportional  to  the  product  of  the  square  of  its  depth  multi- 
plied by  its  width;  consequently,  it  is  more  economical  to 
increase  the  depth  than  the  width. 

TABLE  22. 

CONSTANTS  FOR   TRANSVERSE  STRENGTH. 


Material. 

Constant 
in  Pounds. 

Material. 

Constant 
in  Pounds. 

Metals: 
Cast  Iron 

100 

Woods: 
Birch 

35 

Wrought  Iron  . 

150 

Elm  

25 

Structural  Steel 

1GO 

Ash 

45 

Copper              ... 

50 

Beech 

30 

Brass 

55 

Hickory 

50 

Maple      

GO 

Oak  (American)  . 
Pine  (Pitch)  
Pine  (White).  .    . 

45 
40 
30 

641.     A  cantilever  is  a  beam,  bar,  rod,  etc.,  fixed  at 
one  end  and  subjected  to  a  transverse  stress,  as  shown  in 


262  MECHANICS. 

Fig.    134.     It   has   a   tendency   to   overthrow    the   wall  or 
structure  to  which  it  is  attached. 

The  strength  of  a  cantilever  varies  inversely  as  the  dis- 
tance of  the  load  from  the  point  of  fixing;  and  the  stress 
upon  any  section  varies  directly  as  the  distance  of  the  load 
from  that  section. 

The  strength  of  a  beam,  bar,  rod,  etc.,  which  has  both  its 

ends  supported,  but 
not  fixed,  and  which 
carries  a  load  midway 
between  its  supports, 
is  four  times  that  of  a 
beam  of  the  same 
length,  fixed  into  a 
wall  at  one  end,  and 
carrying  a  load  at  the 
other,  as  in  Fig.  134. 

A  cantilever  uni- 
formly loaded  will  sustain  twice  as  great  a  load  as  one  in 
which  the  load  is  applied  at  the  free  end;  and  a  beam  rest- 
ing on  two  supports  and  uniformly  loaded  will  sustain  twice 
as  great  a  load  as  it  would  if  the  load  were  all  applied  at  the 
middle. 

In  Table  22  is  given  the  safe  transverse  strength  of  bars 
of  different  kinds  of  material,  one  inch  square  and  one  foot 
long,  with  the  load  suspended  from  one  end,  the  other  being 
fixed,  as  shown  in  Fig.  134. 

642.  Rules  and  Formulas  for  the  .Transverse 
Strength  of  Beams: 

Let  a^=  the  depth  of  beam  in  inches; 
w  =  the  width  of  the  beam  in  inches; 
L  —  the  length  of  the  beam  between  its  supports,  in 

feet,  or,  for  cantilever,  the  distance  between  load 

and  fixed  end; 

.S  =  the  safe  transverse  strength,  as  given  in  Table  22; 
W=  the  safe  load  in  pounds. 


MECHANICS. 


263 


For  a  rectangular  or  square  cantilever  to  which  the  load 
is  applied  at  one  end,  as  shown  in  Fig.  134: 

Rule  1 22. —  The  maximum  safe  load  in  pounds  that  should 
be  allowed  at  the  end  of  any  rectangular  or  square  cantilever 
is  equal  to  the  square  of  the  depth  in  inches  multiplied  by  its 
width  in  inches  multiplied  by  the  constant  given  in  the  table, 
and  the  product  divided  by  its  length  in  feet. 

That  is,  W-  d*  w  S 


EXAMPLE.  —  What  is  the  maximum  safe  load  that  can  be  placed  at 
one  end  of  a  cast-iron  bar  which  projects  4  feet,  the  depth  being 
6  inches,  and  the  width  3  inches? 


SOLUTION.—  By  the  rule,  W= 

Substituting  the  values  of  d,  w,  S,  and  Z,  we  have 


6  X  6  X  3  X  100 


=  2,700  pounds.     Ans. 


643.  For  a  cylin- 
drical  cantilever  to 
which  the  load  is  ap- 
plied at  one  end,  as 
shown  in  Fig.  135: 

Rule    123.  —  The 

maximum  safe  load  in 
pounds  that  should  be 
allowed  at  the  end  of 
any  cylindrical  canti- 
lever is  equal  to  the 

cube  of  its  diameter  in  indies  multiplied  by  .6  of  the  constant 
given  in  the  table,  and  the  product  divided  by  its  length  in 
feet. 

That  is,  ' 


EXAMPLE. — What  is  the  maximum  load  that  can  be  placed  with 
safety  at  one  end  of  a  cast-iron  bar  4  inches  in  diameter  that  projects 
3  feet  ? 


204 


MECHANICS. 


SoLUTiON.-By  the  rule,   W  =  . 

Substituting  the  values  of  d,  S,  and  L,  we  have 


4X4X4X.6X100 


=  1,280  pounds.     Ans. 


644.     When  the  load  is  uniformly  distributed  on  a  can- 

tilever of  any  cross-section,  as  shown  in  Fig.   136,  it  will 

sustain  a  load  twice  as 
great  as  when  the  load  is 
applied  at  one  end.  For 
example,  if  the  canti- 
levers in  the  two  exam- 
ples above  were  to  carry 
a  uniformly  distributed 
load,  they  would  sustain 
2,700  X  2  =5,400  pounds, 
and  1,280  X  2  =  2,560 
FIG.  136.  pounds  respectively. 

For  a  rectangular  or  square  beam  the  ends  of  which  merely 

rest  upon  supports, 

and    loaded   in   the 

middle,  as  shown  in 

Fig.   137: 

Rule    124.—  The 

maximum  safe  load 

in  pounds   tJiat  any 

rectangular  or  square 

beam    is   capable  of 

sustaining  at   the  middle,  when    its  ends  merely   rest   upon 

supports,   is  equal  to  four  times  the  square  of  its  depth  in 

inches  multiplied  by   its   width   in  inches  multiplied  by  the 

constant  given   in  the  table,  and  the  product  divided  by  the 

distance  between  its  supports  in  feet. 


That  is, 


FIG.  137. 


EXAMPLE. — What  maximum  safe  load  is  a  bar  of  cast  iron  capable 
of  sustaining  in  the  middle  between  the  supports  on  which  its  ends 


MECHANICS. 


265 


merely  rest,  if  its  depth  is  6  inches,  its  width  3  inches,  and  the  dis- 
tance between  the  supports  is  4  feet  ? 


SOLUTION.—  By  tne  rule,  ^_ 


4X6«x3XlOO 


Ans. 


645.  For  a  cylindrical  beam  supported  at  its  ends  and 
loaded  in  the  middle, 
as  shown  in  Fig.  138: 

Rule   125.—  The 

maximum   safe  load 

in  pounds   that    any 

cylindrical    beam    is 

capable  of  sustaining 

at   the  middle,  when  FIG.  m 

its  ends  merely  rest  upon  supports,  is  equal  to  four  times  the 

cube  of  its  diameter  multiplied  by  .6  of  the  constant  given  in 

the  table,  and  the  product  divided  by  the  distance  between  its 

supports  in  feet. 

...      4</'x.6S 
That  is,  W= j 

EXAMPLE. — What  maximum  safe  load  is  a  bar  of  cast  iron  capable 
of  sustaining  in  the  middle  between  the  supports  on  which  its  ends 
merely  rest,  if  it  is  4  inches  in  diameter,  and  if  the  distance  between 
its  supports  is  3  feet  ? 


SOLUTION. — By  the  rule,  W '  — 


646. 


4x43x.6  X  100 
8 


=  5,1201b.     Ans. 


When  the  load  is  uniformly  distributed  on  a  beam 
of  any  cross-section, 
as  shown  in  Fig.  139, 
it  will  sustain  a  load 
twice  as  great  as 
when  the  load  is  ap- 
plied in  the  middle 

between  the  supports. 

For  example,  if  the  beams  in  the  last  two  examples  were 

to  carry  a  uniformly  distributed  load,  they  would  sustain 

10,800  X  2  —  '21,000  pounds,  and  5,120  X  2  =  10,240  pounds, 

respectively. 


266 


MECHANICS. 


SHEARING,  OR    CUTTING,  STRENGTH    OF 

MATERIALS. 

647.     The   shearing   strength  of    any   material   is   the 
resistance  offered  by  its  fibers  to  being  cut  in  two. 

Thus,  the  pressure  of  the 
cutting    edges    of    an    ordi- 
nary      shearing       machine, 
Fig.   140,   causes  a   shearing 
'MMtMfMMMWMMA  stress  in  the  plane  a  b.      The 
unit  shearing  force   may  be 
^M^MmMmm^  found  by  dividing  the  force 
P  by  the  area  of  the  plane 
ab. 

FiG.~i4o.  648.      Fig.    141  shows   a 

piece  in  double  shear  ;  here  the  central  piece  c  d  is  forced 
out  while  the  ends  i  -  • 

remain    on    their  L     _  _  i  _' 

supports  M  and  N. 

The     shearing 
strength      of     any          w, 
body  is  directly  pro-   i  -  *^ 
portionaltoitsarea.    |_ 

In  the  following 
table  are  given  the  greatest  and  safe   shearing   strengths 
per  square  inch  of  different  kinds  of  materials: 


PIG. 


TABLE  23. 


Material. 

Greatest  Shearing 
Stress  in  Pounds  per 
Square  Inch. 

Safe  Shearing  Stress  in 
Pounds  per  Square  Inch. 

Cast  Iron  

18  000 

1  500  to    3,000 

Wrought  Iron  
Steel  

40,000 
60  000 

4,000  to  10,000 
5,000  to  12,000 

MECHANICS.  2C7 

649.  Formula  for  the  Shearing  Strength  of 
Materials: 

Let  a  =  area  of  cross-section  in  inches; 

5  =  safe  shearing  stress  as  given  in  the  table; 
W  •=•  safe  load  in.  pounds. 

Rule  126. —  The  safe  load  that  any  body  which  is  sub- 
jected to  a  shearing  stress  is  capable  of  sustaining  is  equal  to 
the  area  of  its  cross-section  in  inches  multiplied  by  its  safe 
shearing  stress,  as  given  in  the  table. 

That  is,  IV  =  a  S. 

EXAMPLE.— If  the  beam  in  Fig.  141  is  made  of  wrought  iron,  4 
inches  in  depth  and  2  inches  in  width,  what  steady  shearing  stress  is  it 
capable  of  sustaining  with  safety  ? 

SOLUTION.— Applying  the  rule,  W=  4  x  2  x  10.000  =  80,000  Ib.  This 
result  must  be  multiplied  by  2,  since  the  beam  is  sheared  in  two  places, 
along  the  lines  ec  andfd.  Hence,  the  stress  which  the  beam  will 
safely  sustain  is  80,000  x  2  =  160,000  Ib. 

EXAMPLE. — What  force  is  required  to  punch  a  hole  f  in  diameter 
through  a  steel  plate  £"  thick  ? 

SOLUTION. — It  is  evident  that  punching  is  but  shearing  in  a  circle 
instead  of  a  straight  line.  The  area  punched  (sheared)  is  equal  to 
the  thickness  of  the  plate  multiplied  by  the  circumference  of  a  circle 
having  the  same  diameter  as  the  punched  hole.  For,  if  the  plate  were 
cut  through  one  of  the  diameters  of  the  punched  hole,  and  the  two 
semicircles  were  straightened  out.  the  punched  surface  would  be  a  rect- 
angle, which  would  have  a  length  equal  to  the  circumference  of  a 
circle  whose  diameter  was  equal  to  that  of  the  hole,  and  a  breadth 
equal  to  the  thickness  of  the  plate.  In  this  case,  the  area  =  |  X 
3.1416  X  i  =  .98175  sq.  in.  Table  23  gives  the  ultimate  shearing 
strength  of  steel  as  60,000  Ib.  per  sq.  in.  Hence,  the  total  force  re- 
quired is  .98175  x  60,000  =  58,905  Ib.  Ans. 


EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  greatest  load  that  can  be  safely  carried  by  a  steel 
rectangular  cantilever  at  its  extreme  end.  it  the  bar  is  2*  wide,  3"  deep, 
and  2  ft.  6"  long  ?  Ans.  1. 152  Ib. 

2.  What  is  the  greatest  uniform  load  that  can  be  safely  carried  by 
a  white-pine  girder,  6"  wide,  8"  deep,  16  ft.  long,  and  supported  at  its 
ends?  Ans    5,760  Ib. 


268  MECHANICS. 

3.  A  cast-iron  bar,  If"  in  diameter  and  5  ft.  3"  long,  is  supported  at 
its  ends;  what  load  will  it  safely  sustain  in  the  middle  ?      Ans.  245  Ib. 

4.  What  force  is  required  to  punch  a  !£"  hole  through  a  wrought- 
iron  plate  T7/  thick  ?  Ans.  68,723  Ib. 

5.  What  force  is  required  to  cut  off  the  end  of  a  cast-iron  bar 
whose  diameter  is  2£"  ?  Ans.  88,357  Ib. 


LINE    SHAFTINGS. 

65O.  A  line  of  shafting  is  one  continuous  run,  or  length, 
composed  of  lengths  of  shafts  joined  together  by  couplings. 

The  main  line  of  shafting  is  that  which  receives  the 
power  from  the  engine  or  motor,  and  distributes  it  to  the 
other  lines  of  shafting  or  to  the  various  machines  to  be 
driven. 

Line  shafting  is  supported  by  hangers,  which  are  brackets 
provided  with  bearings,  bolted  either  to  the  walls,  posts, 
ceilings,  or  floors  of  the  building.  Short  lengths  of  shafting, 
called  countershafts,  are  provided  to  effect  changes  of 
speed,  and  to  enable  the  machinery  to  be  stopped  or 
started. 

Shafting  is  usually  made  cylindrically  true,  either  by  a 
special  rolling  process,  when  it  is  known  as  cold-rolled 
shafting,  or  else  it  is  turned  up  in  a  machine  called  a  lathe. 
In  the  latter  case  it  is  called  bright  shafting.  What  is 
known  as  black  shafting  is  simply  bar  iron  rolled  by  the 
ordinary  process,  and  turned  where  it  receives  the  couplings, 
pulleys,  bearings,  etc. 

Bright  turned  shafting  varies  in  diameter  by  £  inch  up 
to  about  3|  inches  in  diameter;  above  this  diameter  the 
shafting  varies  by  |  inch.  The  actual  diameter  of  a  bright 
shaft  is  T^g  of  an  inch  less  than  the  commercial  diameter,  it 
being  designated  from  the  diameter  of  the  ordinary  round 
bar  iron  from  which  it  is  turned.  Thus,  a  length  of  what 
is  called  3-inch  bright  shafting  is  only  2f|  inches  in  diam- 
eter. 

Cold-rolled  shafting  is  designated  by  its  commercial 
diameter ;  thus,  a  length  of  what  is  called  3-inch  shafting 
is  3  inches  in  diameter. 


MECHANICS. 


269 


651.  In  the  following  table  is  given  the  maximum  dis- 
tance between  the  bearings  of  some  continuous  shafts  which 
are  used  for  the  transmission  of  power: 

TABLE    24. 


Diameter 
of  Shaft  in 
Inches. 

Distance  Between  Bearings 
in  Feet. 

Wrought-Iron 
Shaft. 

Steel  Shaft. 

2 

11 

11.5 

3 

13 

13.75 

4 

15 

15.75 

5 

17 

18.25 

6 

19 

20. 

7 

21 

22.25 

8 

23 

24. 

9 

25 

26. 

Pulleys  from  which  considerable  power  is  to  be  taken  should 
always  be  placed  as  close  to  a  bearing  as  possible. 

652.  The  diameters  of  the  different  lengths  of  shafts 
composing  a  line  of  shafting  may  be  proportional  to  the 
quantity  of  power  delivered  by  each  respective  length.  In  this 
connection,  the  positions  of  the  various  pulleys  depend  upon 
the  distance  between  the  pulley  and  the  bearing,  and  upon  the 
amount  of  power  given  off  by  the  pulleys.  Suppose,  for  ex- 
ample, that  a  piece  of  shafting  delivers  a  certain  amount  of 
power,  then,  it  is  obvious  that  the  shaft  will  deflect  or  bend 
less  if  the  pulley  transmitting  that  power  be  placed  close  to  a 
hanger  or  bearing,  than  if  it  be  placed  midway  between  the 
two  hangers  or  bearings.  It  is  impossible  to  give  any  rule 
for  the  proper  distance  of  bearings  which  could  be  used 
universally,  as  in  some  cases  the  requirements  demand  that 
the  bearings  be  nearer  together  than  in  others. 

If  the  work  done  by  a  line  of  shafting  is  distributed  quite 
equally  along  its  entire  length,  and  the  power  can  be  applied 


270 


MECHANICS. 


near  the  middle,  the  strength  of  the  shaft  need  be  only  half 
as  great  as  would  be  required  if  the  power  were  applied  at 
one  end. 

653.     To  compute  the  horsepower  that  can  be  transmit- 
ted by  a  shaft  of  any  given  diameter: 

Let  D  =  diameter  of  shaft; 

R  =  revolutions  per  minute; 
H=  horsepower  transmitted; 
C  =  constant  given  in  Table  25. 

TABLE    25. 

CONSTANTS  FOR    LINE  SHAFTING. 


Material  of  Shaft. 

No  Pulleys 
Between  Bearings. 

Pulleys  Between 
Bearings. 

Steel  or  Cold-Rolled  Iron 
\Vrought  Iron 

65 

70 

85 
95 

Cast  Iron  

90 

120 

In  the  above  table  the  bearings  are  supposed  to  be  spaced 
so  as  to  relieve  the  shaft  of  excessive  bending;  also,  in  the 
third  vertical  column,  an  average  number  and  weight  of 
pulleys,  and  power  given  off,  is  assumed. 

In  determining  the  above  constants,  allowance  has  been 
made  to  insure  the  stiffness  as  well  as  strength  of  the  shaft. 
Cold-rolled  iron  is  considerably  stronger  than  ordinary  turned 
wrought  iron ;  the  increased  strength  being  due  to  the  proc- 
ess of  rolling,  which  seems  to  compress  the  metal  and  so 
make  it  denser,  not  merely  skin  deep,  but  practically  through- 
out the  whole  diameter.  We  have,  then,  the  following: 

Rule  127. —  The  horsepower  that  a  shaft  will  transmit 
equals  the  product  of  the  cube  of  the  diameter  and  the  number 
of  revolutions,  divided  by  the  value  of  C  for  the  given  material. 


That  is. 


H  = 


MECHANICS.  271 

EXAMPLE.— What  horsepower  will  a  3-inch  wrought-iron  shaft 
transmit  which  makes  100  revolutions  per  minute,  there  being  no 
pulleys  between  bearings  ? 

SOLUTION.—  H=  D*  £  R. 
Substituting,  we  have 

3  X  3  X  3  X  100 

H=—   — ^ —        =38. 57  horsepower.     Ans. 

If  there  were  the  usual  amount  of  power  taken  off,  as  mentioned 
above,  we  should  take  C  =  95.  Then,  H=  2?  *  10°  =  28.42  horse- 
power. Ans. 

654.  To  compute  the  number  of  revolutions  a  shaft 
must  make  to  transmit  a  given  horsepower: 

Rule  128. — The  number  of  revolutions  necessary  for  a 
given  horsepower  equals  the  product  of  the  value  of  C  for  the 
given  material  and  the  number  of  horsepower,  divided  by  the 
cube  of  the  diameter.  „ 

That  is,  R  =      *      . 


EXAMPLE.— How  many  revolutions  must  a  3-inch  wrought-iron  shaft 
make  per  minute  to  transmit  28.42  horsepower,  power  being  taken  off 
at  intervals  between  the  bearings  ? 

SOLUTION.—  R  =  ^jf- 

Substituting,  we  have  R  =  ^f  X02?'4<?  =  100  revolutions.     Ans. 

o  X  o  X  « 

655.  To  compute  the  diameter  of  a  shaft  that  will 
tiansmit  a  given  horsepower,  the  number  of  revolutions  the 
shaft  makes  per  minute  being  given: 

Rule  129. —  The  diameter  of  a  shaft  equals  the  cube  root 
of  the  quotient  obtained  by  dividing  the  product  of  the  value 
of  C  for  the  given  material  and  the  number  of  horsepower 
by  the  number  of  revolutions. 

ThatiS) 


EXAMPLE. — What  must  be  the  diameter  of  a  wrought-iron  shaft  to 
transmit  38.57  horsepower,  the  shaft  to  make  100  revolutions  per 
minute,  no  power  being  taken  off  between  bearings  ? 

SOLUTION.—/?  =  y — ^ — . 
Substituting,  we  have 


272  MECHANICS. 

As  the  speed  of  shafting  is  used  as  a  multiplier  in  the  cal- 
culations of  the  horsepower  of  shafts,  a  shaft  having  a  given 
diameter  will  transmit  more  power  in  proportion  as  its  speed 
is  increased.  Thus,  a  shaft  which  is  capable  of  transmitting 
10  horsepower  when  making  100  revolutions  per  minute,  will 
transmit  20  horsepower  when  making  200  revolutions  per 
minute.  We  may,  therefore,  say  the  horsepowers  transmitted 
by  two  shafts  are  directly  proportional  to  the  number  of 
revolutions. 

EXAMPLES   FOR    PRACTICE. 

1.  What  horsepower  will  a  2^"  wrought-iron  shaft  transmit  when 
running  at  110  revolutions  per  minute,  it  being  used  for  transmission 
only  ?  Ans.  24.55  horsepower. 

2.  A  6"  cast-iron  shaft  transmits  150  horsepower;  how  many  revo- 
lutions per  minute  must  it  make,  no  power  being  taken  off  between 
bearings  ?  Ans.  62^  R.  P.  M. 

3.  What  should  be  the  diameter  of  a  wrought-iron  shaft  to  transmit 
100  horsepower  at  150  revolutions  per  minute,  power  being  taken  off 
between  bearings  ?  Ans    4  in.,  nearly. 

4.  The  diameter  of  a  steam-engine  shaft  is  8" ;    what   horsepower 
will  it  transmit,  if  made  of  steel,  when  making  150  revolutions  per 
minute?  Ans.  1,181.54  horsepower. 

5.  The  machines  driven  by  a  certain  line  of  wrought-iron  shafting 
take  their  power  from  various  points  between  the  bearings;  and,  if  all 
were  working  together  at  their  full  capacity,  they  would  require  65 
horsepower  to  drive  them.     What  diameter  should  the  shaft  be  if  it 
runs  at  150  revolutions  per  minute  ?  Ans.  3£  in. 


A  SERIES 

OF 


QUESTIONS   AND    EXAMPLES 

RELATING  TO  THE  SUBJECTS 
TREATED  OF  IN  VOL.    I. 


It  will  be  noticed  that,  although  the  various  questions  are 
numbered  in  sequence  from  1  to  383,  inclusive,  these  ques- 
tions are  divided  into  five  different  sections,  corresponding 
to  the  five  sections  of  the  preceding  pages  of  this  volume. 
Under  the  heading  of  each  section  are  given,  in  parentheses, 
the  numbers  of  those  articles  which  should  be  carefully 
studied  before  attempting  to  answer  any  question  or  to 
solve  any  example  occurring  in  the  section. 


ARITHMETIC. 

(SEE  ARTS.  1  to  174.) 


(1)  What  is  arithmetic  ? 

(2)  What  is  a  number  ? 

(3)  What  is  the  difference  between  a  concrete  number 
and  an  abstract  number  ? 

(4)  Define  notation  and  numeration. 

(5)  Write  each  of  the  following  numbers  in  words: 
980;  605;  28,284;  9,006,042;  850,317,002;  700,004. 

(6)  Represent  in  figures  the  following  expressions: 
Seven  thousand,  six  hundred.      Eighty-one  thousand,  four 

hundred  two.  Five  million,  four  thousand,  seven.  One 
hundred  eight  million,  ten  thousand,  one.  Eighteen  million, 
six.  Thirty  thousand,  ten. 

(7)  What  is  a  fraction  ? 

(8)  What  are  the  terms  of  a  fraction  ? 

(9)  What  does  the  denominator  show  ? 

(10)  What  does  the  numerator  show  ? 

(11)  How  do  you  find  the  value  of  a  fraction  ? 

(12)  Is  J¥3-  a  proper  or  an  improper  fraction   and  why  ? 

(13)  Write  three  mixed  numbers. 

(14)  Reduce  the  following  fractions  to  their  lowest  terms : 

I,  A,  A,  if  Ans-  *.  *.  *.  *• 

(15)  Reduce   6  to  an  improper  fraction  whose  denomi- 
nator is  4.  Ans.  ^. 

(16)  Reduce  7|,  13-^,  and  10J  to  improper  fractions. 

Ans.      ,  . 


278  ARITHMETIC. 

and   17  thousandths,   and    the  sum   of  53  hundredths  and 
274  thousandths  ?  f  (#)     008 

8  :U 

(d)  .786967. 

(42)  It  is  desired  to  increase  the  capacity  of  an  electric- 
light  plant  to  1,500  horsepower  by  adding  a  new  engine.     If 
the  indicated   horsepower  of  the  engines  already  in  use  is 
482f,   316£,  and  390|,  what    power    must  the    new    engine 
develop?  Ans.  310.  llf  H.  P. 

(43)  On  a  certain  morning  7,240  gallons  of  water  were 
drawn  from  an  engine-room  tank,  and  4,780  gallons  were 
pumped  in.     In  the  afternoon  7,633  gallons  were  drawn  out, 
and  8,675  gallons  pumped  in.      How  many  gallons  remained 
in  the   tank  at  night,   if  it  contained  3,040  gallons  at  the 
beginning  of  the  day?  Ans.  1,622  gal. 

(44)  A  metal  stack  45  feet  high  is  made  up  of  7  plates, 
6  of  which  are  7  feet  long.    Allowing  15  inches  for  laps,  what 
is  the  length  of  the  seventh  plate?  Ans.  4  ft.  3  in. 

(45)  The  inside  diameter  of  a  6-inch  steam  pipe  is  6.06 
inches,  and  the  outside  diameter  is  6.62  inches.      How  thick 
is  the  pipe?  Ans.  .28  in. 

(46)  Find  the  products  of  the  following: 

(a)  526,387  X  7;  (b)  700,298  X  17;  (c)  217  X  103  X  67. 

f  (a)  3,684,709. 
Ans.  ]  (&)  11,905,066. 
(  (c)  1,497,517. 

(47)  If  your  watch  ticks  once  every  second,  how  many 
times  will  it  tick  in  one  week?  Ans.  604,800. 

(48)  Find  the  products  of  the  following  expressions : 

(a)    .013  X  .107;  (b)  203  X  2.03  X  .203  ;  (c)  (2.7  X  31.85) 
X  (3.16 -.316);  (d)  (107.8  +  6.541  -  31.96)  X  1.742. 

(a)  .001391. 


.   (b)  83.65427. 
'   (c]   244.56978. 
(d\  143.507702. 


ARITHMETIC.  279 

(49)  An  engine  and  boiler  in  a  manufactory  are  worth 
$3,246.     The  building  is  worth  three  times  ,as  much  plus 
$1,200,  and  the  tools  are  worth  twice  as  much  as  the  building 
plus  $1,875;  (a)  what  is  the  value  of  the  building  and  tools? 
(b}  What  is  the  value  of  the  whole  plant? 

Ans.  \  <*)  *34>G89' 
(  (b)  $37,935. 

(50)  How  many  square  feet  of  heating  surface  are  in  the 
tubes  of  a  boiler  having  GO  3-inch  tubes,  each  15£  feet  long, 
if  the  heating  surface  of  each  tube  per  foot  in  length  is  .728 
square  foot?  Ans.  677.04  sq.  ft. 

(51)  Suppose  that  in  one  hour  10  pounds  of  coal  are  burned 
per  square  foot  of  grate  area  in  a  certain  boiler,  and  that 
9  pounds  of  water  are  evaporated  per  pound  of  coal  burned. 
If  the  grate  area  is  30  square  feet,  how  many  pounds  of 
water  would  be  evaporated  in  a  day  of  10  hours? 

Ans.  27,000  Ib. 

(52)  How  many  feet  does  the  piston  of  a  steam  engine 
pass  over  in  a  week  of  6  days,  running  8£  hours  a  day,  if  the 
length  of  the  stroke  of  the  engine  is  1^-  feet,  and  the  number 
of  revolutions  per  minute  160?  Ans.  1,468,800  ft. 

(53)  A  number  of  boilers  are  constantly  fed  by  3  pumps; 
the  first  delivers  1J  gallons  per  stroke,  and  runs  at  75  strokes 
per  minute;  the  second  delivers  $•  of  a  gallon  per  stroke,  and 
runs  at  115  strokes  per  minute;  the  third  delivers  If  gallons 
per  stroke,  and  runs  at  96  strokes  per  minute.      How  many 
gallons  of  water  are  fed  to  the  boilers  per  hour? 

Ans.  21,022.5  gal. 

NOTE.— If  in  the  following  examples  there  be  a  remainder,  carry  the 
quotient  to  four  decimal  places. 

(54)  Divide  the  following: 

(a)  962,842  by  84;    (b)  39,728  by  63;    (c)  29,714  by  108; 
(d)  406,089  by  135.  {  (a)    11,462.4048. 


;   (b)    630.6032. 
"   (c)   275.1296. 
(d)  3,008.0667. 


280 


ARITHMETIC. 


(55)     Solve  the  following: 
(a)  35-  A;    (b)  ^-5-3;    (c) 


•¥•-*•  9; 

(<*)  W-*-  A;  (') 

Ans.  - 

(a)    112. 
(*)    A- 

(0  H- 

375  X* 

(  (a)  1.75. 
Ans.  ]  (*)   1.75. 
(   (c)     .5. 

T5ir  -  •  125' 

(56)     Solve  the  following: 
(a)  .875-|;  (b)  *-s-.5;  ' (c) 


(57)     Solve  the  following  by  cancelation  : 

(a)  (72  X  48  X  28  X  5)  -  (84  X  15  X  7  X  6). 

(b)  (80  X  60  x  50  X  16  X  14)  —  (70  X50  X  24  x  20). 

Ans 


(b)  32. 


(58)     Find  the  values  of  the  following  expressions: 


la)        -  (M        -   (,) 

'    f  '   l 


1.25X20X3 

87  +  88      ' 


(a)  37*. 
Ans.      (b)  .75. 
(c)  210f 


459  +  32 

(59)  The  distance  around  a  cylindrical  boiler  is  166.85 
inches.      If  there  are  72  rivets  in  one  of  the  circular  seams, 
find  what  the   pitch  (distance  between  the  centers  of  any 
two  rivets)  of  the  rivets  is.  Ans.  2.317  -+-in. 

(60)  A   keg   of  |-  x  2|  inches  boiler  rivets   weighs    100 
pounds,  and  contains   133   rivets.     What  is  the   weight   of 
each  rivet  ?  Ans.  .75  +  Ib. 

(61)  The  distance  around  a  wheel  equals  3.1416    times 
its   diameter,   or  the   distance    across    it.     If   the    distance 
around  a  fly-wheel  is  56.5488  feet,  what  is  the  diameter  of 
a  wheel  half  as  large  ?  Ans.  9  ft. 

(62)  If  980,000  bricks  are   required  to  build  an  engine 
house,    how    many    days    will  it  take  for  6   teams  to  draw 
them,  each  team  drawing  8  loads  a  day,  and  there  being 
1,600  bricks  to  a  load  ?  Ans.  12.76  +  days. 


ARITHMETIC.  281 

(63)  If  a  mechanic  earns  $1,500  a  year  for  his  labor,  and 
his  expenses  are   $968   per  year,  in  what  time  can  he  save 
enough  to  buy  28  acres  of  land  at  $133  an  acre? 

Ans.  7  years. 

(64)  The  numerator  of  a  fraction  is  28,  and  the  value 
of  the  fraction  £;  what  is  the  denominator  ?  Ans.  32. 

(65)  From  1  plus  .001  take  .01  plus  .000001. 

Ans.   .990999. 

(66)  A  freight    train  ran    305    miles    in   one  week,  and 
3  times  as  far,   lacking  246  miles,  the  next  week;  how  far 
did  it  run  the  second  week  ?  Ans.  849  miles. 

(6-7)  If  the  driving  wheel  of  a  locomotive  is  16  ft.  in 
circumference,  how  many  revolutions  will  it  make  in 
going  from  Philadelphia  to  Pittsburg,  the  distance  of 
which  is  354  miles,  there  being  5,280  feet  in  one  mile  ? 

Ans.   116,820  rev. 

(68)  How  many  inches  in  .875  of  a  foot  ?        Ans.  10£  in. 

(69)  What  decimal  part  of  a  foot  is  T3T  of  an  inch  ? 

Ans.  .015625  ft. 

(70)  If  water  be    conducted    from  a  tank  by  7  lengths 
of  gas  pipe  coupled  together,    each    length   being    12    feet 
6  inches  long,  £  of  an  inch  being  added  at  each  of  the  joints 
for  coupling  together,  how  far  from  the  tank  is  the  water 
discharged  ?     The    first  length    screws    into  the  tank  £  of 
an  inch.  Ans.  87.8125  ft. 

(71)  If  by  selling    a    carload    of    coal    for   $82.50,   at    a 
profit  of  $1.65  per  ton,  I  make  enough  to  pay  for  72.6  ft. 
of  fencing  at  $.50  a  foot,  how  many  tons  of  coal  were  there 
in  the  car  ?  Ans.   22  tons. 

(72)  The   connection   between  an  engine  and   boiler   is 
made  up  of  6  lengths  of  pipe,   three  of  which  are  14  feet 
5  inches  long,   two  12  feet  6  inches  long,   and  one  8  feet 
10  inches  long.     If  the  pipe  weighs  10£  pounds  per  foot,  what 
is  the  total  weight  of  the  pipe  used  ?  Ans.  809.375  Ib. 

(73)  Four   bolts  are   required,  2£,  6£,  3^,  and  4  inches 
long.      How    long   a    piece  of    iron   will   be   required   from 


282  ARITHMETIC. 

which  to  cut  them,  allowing  T\  of  an  inch  to  each  bolt  for 
cutting  off  and  finishing  ?  Ans.  18T3g-. 

(74)  A  double  belt  of  a  certain  width  can  transmit  64£ 
horsepower.      How  many  horsepower  can  two  single  belts  of 
the  same  width  transmit,  when  running  under  the  same  con- 
ditions, supposing  that  the  double  belt  is  capable  of  trans- 
mitting y  as  much  power  as  one  of  the  single  belts  ? 

Ans.   90.3  H.  P. 

(75)  The  lengths  of  belting   required   to   connect  four 
countershafts  with  the  main  line  shaft  were  18  feet  6  inches, 
16  feet  9£  inches,  22  feet  2  inches,  and  20  feet  8£  inches.    How 
many  feet  of  belting  were  required  ?  Ans.   78£  ft. 

(76)  In  a  steam-engine  test  of  an  hour's  duration,  the 
horsepower  developed  was  found   to   be   as  follows,   at  10- 
minute   intervals:    48.63,    45.7,    46.32,    47.9,   48.74,    48.38, 
48.59.     What  was  the  average  horsepower  ? 

Ans.  47.75+  H.  P. 


ARITHMETIC. 

(SEE  ARTS.  175-339.) 


(77)  What  is  25$  of  8,428  Ib.  ?  Ans.  2,107  Ib. 

(78)  What  is  1$  of  $100  ?  Ans.  $1. 

(79)  What  is  \%  of  $35,000  ?  Ans.   $175. 

(80)  What  per  cent,  of  50  is  2  ?  Ans.  4$. 

(81)  What  per  cent,  of  10  is  10  ?  Ans.   100$. 

(82)  Solve  the  following: 

(a)  Base  — $2,522,  and  percentage  =  $176.54.  What  is 
the  rate  ? 

(b}  Percentage^  16.96,  and  rate  =  8$.  What  is  the 
base? 

(c)  Amount  =216.7025,  and  base  =  213.5.     What  is  the 
rate  ? 

(d)  Difference  =  201.825,  and  base  =  207. 


(83)  The  coal  consumption  of  a  steam  plant  is  5,500  Ib. 
per  day  when  the  condenser  is  not  running,  or  an  increase 
of  15$  over  the  consumption  when  the  condenser  is  used. 
How  many  pounds  are  used  per  day  when  the  condenser  is 
running?  Ans.  4,782.61  Ib. 

(84)  An  engineer  receives  a  salary  of  $950.     He  pays  24$ 
of  it  for  board,  12£$  of  it  for  clothing,  and  17$  of  it  for  other 
expenses.      How  much  does  he  save  a  year?      Ans.   $441.75. 

(85)  If  37|$  of  a  number  is  961.38,  what  is  the  number? 

Ans.  2,563. 6a 


284  ARITHMETIC. 

(86)  A  man  owns  f  of  a  manufacturing  plant.   30#  of  his 
share  is  worth  $1,125.     What  is  the  whole  property  worth  ? 

Ans.   $5,000. 

(87)  What  number  diminished   by  35$  of  itself  equals 
4,810?  Ans.   7,400. 

(88)  The    volume    of   the    clearance  in   a  steam-engine 
cylinder  is  found  to  be  18.3cu.   in.,  and  the  volume  of  the 
cylinder,  neglecting  the  clearance,  254.5  cu.  in.     What  per- 
centage of  the  cylinder  volume  is  the  clearance  ? 

Ans.   7.2$,  nearly. 

(89)  The  distance  between  two  stations  on  a  certain  rail- 
road is  16.5  miles,  which  is  12£$  of  the  entire  length  of  the 
road.     What  is  the  length  of  the  road  ?          Ans.   132  miles. 

(90)  The  speed  of  an  engine  running  unloaded  was  1|$ 
greater  than  when  running  loaded.      If  it  made  298  revolu- 
tions per  minute  with  the  load,  what  was  its  speed  running 
unloaded  ?  Ans.   302.47  revolutions  per  minute. 

(91)  Reduce  4  yd.  2  ft.  10  in.  to  inches.          Ans.  178  in, 

(92)  Reduce  3,722  in.  to  higher  denominations. 

Ans.  103  yd.  1  ft.  2  in. 

(93)  How  many  seconds  in  5  weeks  and  3.5  days  ? 

Ans.  3,326,400  sec. 

(94)  Reduce  764,325  cu.  in.  to  cu.  yd. 

Ans.  16  cu.  yd.  10  cu.  ft.  549  cu.  in. 

(95)  How  many  gallons  of  water  can  be  put  into  a  tank 
holding  4  bbl.  10  gal.  3  qt.  ?  Ans.  136|  gal. 

(96)  A  carload  of  coal  weighed  16  T.  8  cwt.  75  Ib.     How 
many  pounds  did  this  amount  to  ?  Ans.  32,875  Ib. 

(97)  Reduce  25,396  Ib.  to  higher  denominations. 

Ans.  12  T.  13  cwt.  96  Ib. 

(98)  Reduce  25,396  pt.  to  higher  denominations. 

Ans.  100  bbl.  24  gal.  2  qt. 

(99)  What  is  the  sum  of  2  yd.  2  ft.  3  in. ;  4  yd.  1  ft.  9  in. ; 
2  ft.  7  in.  ?  Ans.  8  yd.  7  in. 


ARITHMETIC.  285 

(100)  What  is  the  sum  of  3  gal.  3  qt.  1  pt. ;  6  gal.  1  pt. ; 
4  gal.  8  qt.  5  pt.  ?  Ans.  16  gal.  2  qt.  1  pt. 

(101)  From  52  yd.  2  ft.  9  in.  take  115  ft. 

Ans.  14  yd.  1  ft.  9  in. 

(102)  From   a   barrel  of  machine  oil  is  sold  at  one  time 
10  gal.  2  qt.  1  pt.,  and  at  another  time  16  gal.  3  qt.      How 
much  remained  ?  Ans.  4  gal.  1  pt. 

(103)  If  1  iron  rail  is  17  ft.  3  in.  long,  how  long  would 
51  such  rails  be,  if  placed  end  to  end  ?         Ans.  879  ft.  9  in. 

(104)  Multiply  3  qt.  1  pt.  by  4.7.  Ans.  32.9  pt. 

(105)  How  many  iron  rails,  each  30  ft.  long,  are  required 
to  lay  a  single  railroad  track  23  miles  long  ? 

Ans.  8,096  rails. 

(106)  The   main   line   shaft   of   a   mill   is   composed  of 
4  lengths  each  15  ft.  5  in.  long,  of  one  piece  14  ft.  8  in.  long, 
and  one  piece   8  ft.  10  in.  long.      If  there  are   6   hangers 
spaced   equally   distant   apart,   one   being   placed   at   each 
extremity  of  the  shaft,  what  is  the  distance  between  the 
hangers  ?  Ans.  17  ft.  f  in. 

(107)  The    distance    around    a   wheel    is   approximately 
2Tj?  times  the  diameter.      If  the  diameter  of  a  fly-wheel  is 
9  ft.  6£  in.,  find  the  distance  around  it.      Ans.  29  ft.  llf  in. 

(108)  If  the  length  of  a  boiler  shell  is  18  ft.  ll^  in.,  how 
many  rivets  should   there    be   in  one    of   the    longitudinal 
seams  if  it  is  a  single-riveted  seam,  supposing  the  rivets  to- 
have  a  pitch  of  1^  in.,  and  the  two  end  rivets  to  be  1|  in. 
from  each  end  of  the  boiler  ?  Ans.  181  rivets. 

(109) 

(a)  What  is  the  second  power  of  108  ? 

(b)  What  is  the  third  power  of  181.25  ? 

(c)  What  is  the  fourth  power  of  27.61  ? 

(a)  11,664. 

Ans.  ]   (&)  5,954,345.703125. 
(c)   581,119.73780641. 


286 


ARITHMETIC. 


(110)     Solve  the  following  : 
(a)  106s;    (b)    (182£)';    (c)  .005°; 
(/)  67.85';  (g)  967,845s;  (k)  (A)' 


Ans.  -I 


(d)  .0063';    (e)  10.06s; 
(0  (i)'. 

(«)    11,236. 

(£)    33,169.515625. 

(c)  .000025. 

(d)  .00003969. 

(e)  101.2036. 
(/)  4,603.6225. 

(g)    936,7^3,944,025. 


Ans. 


(111)  Solve  the  following : 

(a)  7533;    (b)    9S7.43;    (c)    .0053;    (d)  .40443;    (e)    .0133s; 
(/)  301.011s;  (g)  (i)s;  (A)  (3|)3. 

f  («)    426,957,777. 

(b)  962,674,279.624. 

(c)  .000000125. 

(d)  .066135317184. 

(e)  .000002352637. 

(/)  27,273,890.942264331. 

(g}  rnr- 
L  (//)    52.734375. 

(112)  What  is  the  fifth  power  of  2  ?  Ans.  32. 

(113)  What  is  the  fourth  power  of  3  ?  Ans.  81. 

(114)  What  is  the  ninth  power  of  7  ?  Ans.  40,353,607. 

(115)  Solve  the  following: 

(a)  1.2';  (b)  II5;  (c)  1';  (d)  .01';  (e)  .l\ 

(a)  2.0736. 

(b)  161,051. 
AnsJ    (c)  1. 

(d)  .00000001. 

(e)  .00001. 

(116)  In    what     respect    does     evolution     differ    from 
involution  ? 


ARITHMETIC 


287 


(117)  Find  the  square  root  of  the  following: 

(a)  3,486,784.401;  (/>)  9,000,099.4009;  (r)  .001225; 
(d)  10,795.21;  (e)  73,008,05;  (/)  9;  (g)  .9. 

(a)  1,8G7.29-|-. 

(b)  3,000.016. 

(c)  .035. 
Ans.  {    (d}  103.9. 

(e)    270.2. 
(/)  3. 
(g)   .948+. 

(118)  What  is  the  cube  root  of  the  following: 

(a)  .32768?  (b)  74,088?  (c)  92,416?   (d)  .373248  ? 


(e)  1,758.416743?  (/)  1,191,016?     (g) 


Ans.  - 


(119) 


('<)  &V  ? 

(a)  .689+. 

(b)  42. 

(c)  45.2115+. 

(d)  .72. 

(e)  12.07. 
(/)  106. 

(e)  •«• 

(A)  .472+. 


Find  the  cube  root  of  2  to  four  decimal  places. 

Ans.   1.2599+. 


(120)  Find  the  cube  root  of  3  to  three  decimal  places. 

Ans.  1.442+. 

(121)  Solve  the  following: 

(a)  i/123.21; 


Ans. 


(a)  11.1. 

(b)  10.72+. 
}    (c)    709. 

[  (d)   .0203. 


288 


ARITHMETIC. 


(122)     Solve  the  following : 


(a)  I/.0065;     (b)  y7021;     (c}  f/8,  036,  054,  027; 

(d)  ^.000004096;     (e)  #T7. 
|(«)    .186  + 
(£)    .27+. 
(r)   2,003. 
(</)  .016. 
(f)   2.57+. 
(123)     Extract  the  square  root  of: 

(£)    .3364.                                            (£)    .58. 
(c)    .1.                                    Ans.-     (V)    .31622+, 
(d)   25.  Of.                                            (</)   5.00749. 
(e)    .OOOf                                              (r)    .02108. 

(124) 

Find  the  value  of  x  in  the  following: 

11.7  :  13::  20  :  x. 

Ans.  22.22+. 

(125) 

(a)  20  +  7  :  10  +  8  : 

3  :  x.         A        I  (a)  2. 

(b)  122  :  1002::4  :  x. 

M  (J)   277.7+. 

(126) 

wl-ri!<*):a= 

8      ,  ,    2         x       '       15      60 
F6;  (C>IO=100;  (^46  =  ^; 

/  \    10  _ 

x 

|(«)   *  ^  12. 

^'  1~50~ 

600* 

»>»«; 

(^  )   *  =  20. 

(d)  x  =  180. 

(,)   *-40. 

(127) 

x  :  5  ::  27  :  12.5. 

Ans.   10|. 

(128)   . 

45  :  60  ::  x  :  24. 

Ans.   18. 

(129) 

x  :  35  ::  4  :  7. 

Ans.   20. 

(130) 

9  :  x  ::  6  :  24. 

Ans.   36. 

(131) 

fl^OOO  :  vT^3l::27 

:  x.                             Ans.  29.7. 

(132) 

64  :  81  ::  21'  :  x\ 

Ans.   23.625. 

(133) 

7  +  8  :  7  ::  30  :  x. 

Ans.   14. 

(134) 

If  a  piece  of  2-inch 

shafting  3£  ft.    long   weighs 

37.45  lb., 

how  much  would  a 

piece  6f  ft.  long  weigh  ? 

Ans.   72.225  lb. 


ARITHMETIC.  289 

(135)  The  intensity  of  heat  from  a  burning  body  varies 
inversely  as  the  square  of  the  distance  from  it.     If  a  ther- 
mometer held  G  ft.  from  a  stove  rises  24  degrees,  how  many 
degrees  will  it  rise  if  held  12  ft.  from  the  stove  ?       Ans.   0°. 

(136)  If  sound  travels  at  the  rate  of  6. 100  ft.  in  5£  sec., 
how  far  does  it  travel  in  1  min.  ?  Ans.   07,200  ft. 

(137)  If  a  railway  train  runs  444  mi.  in  8  hr.  40  min., 
in  what  time  can  it  run  1,060  mi.  at  the  same  rate  of  speed  ? 

Ans.   20  hr.  41.44  min. 

(138)  If  a  pump  discharging  135   gal.  per  minute  fills  a 
tank   in    38    minutes,    how   long    would    it    take    a    pump 
discharging  85  gal.  per  minute  to  fill  it  ?         Ans.   60T6T  min. 

(139)  If  a  quantity  of  babbitt  metal  contains    8    Ib.   of 
copper,  8  Ib.  of  antimony,  and  80  Ib.  of  tin,  how  much  copper 
will  32  Ib.  of  the  same  metal  contain  ?  Ans.   2f  Ib. 

(140)  The  distances  around    the    drive    wheels   of  two 
locomotives  are  12.50  ft.  and   15.7  ft.,   respectively.      How 
many  times  will  the  larger  turn  while  the  smaller   turns 
520  times  ?  Ans.  410  times. 

(141)  If  a  cistern   28  ft.   long,  12  ft.   wide,  10  ft.   deep 
holds   510  bbl.   of  water,   how  many  barrels  of  water  will 
a  cistern  hold  that  is  20  ft.  long,  17  ft.  wide,  and  0  ft.  deep  ? 

Ans.   309T9¥  bbl. 


MENSURATION  AND  USE  OF  LETTERS 

IN 

ALGEBRAIC  FORMULAS. 

(SEE  ARTS.  340-432.) 


A  =  5  //  =  200 

B  =  10  x  =  12 

z=3.5          Z>=120 

Work  out  the  solutions  to  the  following  formulas,  using 
the  above  values  for  the  letters : 

(142)  C=D~X..  Ans.   C=S. 

(143)  6=17'  +  f+/>.  An,   C=l«i. 

2;r  +  6 

3'24(iS/<  Ans.  r  =  187.«a9+. 


7^+T5- 

(U0>     "  =  ^.000184^-^  An,  »  =  «.35+. 

10j^~^)3.  Ans.  /=  12,800. 

Ans.  g  =  5. 
Ans.  k  =7.071+. 

yrx 


292  MENSURATION. 


(150)  T=         —  — .  Ans.    7- =10. 

V      h+!L(A*-BY 

(151)  If  one  of  the  angles  formed  by  one  straight  line 
meeting  another  straight  line  equals  152°  3',  what  is  the  other 
angle  equal  to?  Ans.  27°  57'. 

(152)  How  many  seconds  are  there  in  140°  17'  10". 

Ans.  505,030". 

(153)  Write  a  definition  for  a  degree,  as  applied  to  the 
measurement  of  angles. 

(154)  (a)  How  many  degrees  are  there  in  240  minutes  ? 
(b)  How  many  seconds  ?  (  (a)  4°. 

3"  \  (J)    14,400". 

(155)  Draw  an  obtuse  angle,  a  right  angle,  and  an  acute 
angle.      State  the  name  of  each  angle  by  using   letters  to 
designate  them. 

(150)  Draw  a  rhombus  and  then  draw  a  rectangle  having 
the  same  area. 

(157)  Can  a   quadrilateral  be  formed  with  lines  whose 
lengths   are  20  inches,  9    inches,    4   inches,  and  7    inches  ? 
Give  reasons  ? 

(158)  A  sheet   of  zinc  measures  11£  inches  by  2^   feet. 
How  many  square  inches  does  it  contain?      Ans.  345  sq.  in. 

(159)  If  the  zinc   in  the  last  example  weighs  5£  pounds, 
what  is  its  weight  per  square  foot?  Ans.  2.19  Ib. 

(1GO)  How  many  boards  1C  feet  long  and  5  inches  wide 
would  be  required  to  lay  a  floor  measuring  15  X  24  feet? 

Ans.  54  boards. 

(161)  A  lot  of  land  is  in  the  shape  of  a  trapezoid.  It  is 
16  rods  long,  9  rods  wide  at  the  front,  and  6  rods  wide  at  the 
back.  The  front  and  back  being  parallel,  what  part  of  an 
acre  does  the  lot  contain?  Ans.  £  of  an  acre. 


MENSURATION.  293 

(163)     The  accompanying  figure  shows  the  floor  plan  of 


Switch 


an  electric-light  station.     From  the  dimensions  given,  cal- 
culate the  number  of  square  feet  of  unoccupied  floor  space. 

Ans.   2,059.08  sq.  ft. 

(163)  A   sidewalk  10  feet  wide  extends   around    a  city 
block  528  feet  long  and  352  feet  wide.     Assuming  that  there 
is  no  space  between  the  walk  and  the  buildings,  how  many 
square  yards  does  the  walk  contain?  Ans.  2,000  sq.  yd. 

(164)  How  many  square  yards  of  plastering  will  be  re- 
quired for  the  four  side  walls  of  a  hall  90  feet  long,  50  feet 
wide,  and  20  feet  high,   with  four  doors  5£  X  10  feet  and 
fourteen    windows   5  X  11    feet?      There   is   a  baseboard   9 
inches  high.  Ans.  490.72  sq.  yd. 

(165)  A  triangle  has  three  equal  angles  ;  what  is  it  called  ? 

(166)  If  a  triangle  has  two  equal  angles,  what  kind  of  a 
triangle  is  it  ? 

(167)  Can  a  triangle  be  formed  with  three  lines  whose 
lengths  are  12  inches,  7  inches,  and  4  inches  ?     Give  reasons 
for  your  opinion. 


294  MENSURATION. 

(168)  (a)  What  is  the  altitude  of  an  equilateral  triangle 
whose  sides  are  each  G  feet  ?     (b}  What  is  its  area  ? 

(  (a)  5.196ft. 
5"  (  (b)  15.588  sq.  ft. 

(169)  In  a  triangle  A  B  C,  angle  A  =  23°,  and  B  =  32°  32' ; 
what  does  angle  C  equal  ?  Ans.  C  =  124°  28'. 

(170)    In    the    figure,    if    A    D=W 
inches,  A  B—  24  inches,  and  B  C=  13£ 
inches,  how  long  is  D  £,  D  R  being  par- 
allel to  B  Cl          Ans.  DE=  5.625  in. 
(171)     An    engine    room    is    52   feet 
long  and  39  feet  wide.     How  many  feet 
is  it  from  one  corner  to  a  diagonally  op- 
posite one,  measured  in  a  straight  line? 
Ans.  65  ft. 
(172)     A   ladder   24   feet   long    rests 

against  a  house  with  its  upper  end  8  feet  from  the  ground. 
How  far  on  the  ground  is  the  lower  end  of  the  ladder  from 
the  house  ?  Ans.  22  ft.  7£  +  in. 

(173)  (a)  Show  why  it  is  that  the  area  of   a  triangle 
equals  one-half  the  product  of  the  base  by  the  altitude,      (b) 
Does  it  make  any  difference  which  side  is  taken  as  the  base  ? 

(174)  (a)  The  area  of  an  isosceles  triangle  is  200  square 
inches.     If  its  altitude  is  20  inches,  how  long  is  its  base  ?    (/;) 
What  is  the  length  of  one  of  its  equal  sides  ? 

j  (a)  20  in. 
5'  I  (b)  22.36  in. 

(175)  In  an  equilateral  heptagon  one  of  the  sides  equals 
3  inches;  what  is  the  length  of  the  perimeter  ?      Ans.  21  in. 

(176)  The  perimeter  of  a  regular  decagon  is  40  inches; 
what  is  the  length  of  the  side  ?  Ans.  4  in. 

(177)  What  is  one  angle  of  a  regular  dodecagon  equal  to  ? 

Ans.  150°. 

(178)  The  area  of  a  regular  pentagon  is  43  square  inches. 
If  one  side  is  5  inches  long,  what  is  the  perpendicular  dis- 
tance from  the  center  to  one  side  ?  Ans.  3.44  in. 


MENSURATION. 


205 


(179)  It  is  required  to  make  a  miter-box,  in  which  to  cut 
molding  to  fit  around  an  octagon  post.     At  what  angle  with 
the  side  of  the  box  should  the  saw  run  ?  Ans. 

(180)  Calculate  the  area 
of  the  irregular  polygon, 
Fig.    3.     The    dimensions 
are  to  be  obtained  by  meas- 
uring.    Ans.  1.78  +  sq.  in. 

(181)  An  angle  inscrib- 
ed   in   a  circle    intercepts 
one-fourth  the   circumfer- 
ence.    How  many  degrees 
are  there  in  the  angle  ? 

Ans.  45°. 

(182)  If  the  distance  between  two  opposite  corners  of  a 
hexagonal  nut  is  two  inches,  what  is  the  distance  between 
two  opposite  sides  ?  Ans.  1.732  +  in. 

(183)  In  the  accompanying  figure,  if 
the  distance  B  I  is  G  inches  and  H  K  18 
inches,  what  is  the  diameter  of  the  circle  ? 

B  Ans.  19.5  in. 

(184)  In    the    same    figure,     if    the 
diameter  A  B  =  32£  feet,  and  the  dis- 
tance /  B  —  8  feet,  what  is  the  length 

Ans.  28  ft. 


FIG.  4. 
of  the  chord  H  Kl 


(185)     The  trunk  of  a  tree  measures  7.854  feet  around  it; 
what  is  its  diameter  ?  Ans.  2£  ft. 

(180)     How  many  revolutions  will  a  72-inch  locomotive 
driver  make  in  going  one  mile  ?       Ans.  280.112  revolutions. 

(187)  A  pipe  has  an   internal  diameter  of  6.06   inches; 
what  is  the  area  of  a  circle  having  this  diameter  ? 

Ans.  28.8427  sq.  in. 

(188)  The  area  of  a  circle  having  a  diameter  correspond- 
ing to  the  internal  diameter  of  a  certain  pipe  is  113.0973 
square  inches.     What  is  the  outside  diameter,  the  pipe  being 
f  inch  thick  ?  Ans.  13  J  inches. 


296  MENSURATION. 

(189)  How  long  must  the  arc  of  a  circle  be  to  contain 
12°,  supposing  the  radius  of  the  circle  to  be  G  inches  ? 

Ans.  1.25664  in. 

(190)  Calculate  the  area  of  a  flat  circular  ring  whose  out- 
side diameter  is  22  inches,  and  whose  inside  diameter  is  21 
inches.  Ans.  33.7722  sq.  in. 

(191)  Find  the  area  of  a  segment  of  a  circle  whose  diam- 
eter is  56  inches,  the  height  of  the  segment  being  5  inches. 

Ans.  108.5  sq.  in. 

(192)  What  are  the  dimensions  of  the  end  of  the  largest 
square  bar  that  can  be  planed  from  an  iron  bar  2  inches  in 
diameter  ?  Ans.  1.4142  in.  square. 

(193)  What  is  the  area  of  the  sector  of  a  circle  15  inches 
in  diameter,  the  angle  between  the  two  radii  forming  the 
sector  being  12£°  ?  Ans.   6.1359  sq.  in. 

(194)  (a)  What    would    be  the  length  of  the   side  of  a 
square  metal  plate  having  an  area  of  103.8691  square  inches  ? 
(b]  What  would   be  the  diameter  of  a  round  plate  having 
this  area  ?     (c )  How  much  shorter  is  the  circumference  of 
the  round  plate  than  the  perimeter  of  the  square  plate .? 

!(a)  10. 1916  in. 
(b)  11£  in. 
(c)  4. 638  in. 

(195)  A  plate  46  inches  long  is  to  be  rolled  into  a  shell; 
what  will  be  the  diameter  of  the  shell  if  5  inches  are  allowed 
for  lap?  Ans.    13.05  in. 

(196)  What  is  the  convex  area  in  square  feet  of  a  section 
of  a  smokestack  26  inches  in  diameter  and  10£  feet  long  ? 

Ans.   71. 4714  sq.  ft. 

(197)  Find  the  area  in  square  feet  of  the  entire  surface 
of  a  hexagonal  column  12  feet  long,  each  edge  of  the  ends 
of  the  column  being  4  inches  long  ?          Ans.   24.6774  sq.  ft. 

(198)  Find  the  cubical  contents  of  the  above  column  in 
cubic  inches.  Ans.   5,985.9648  cu.  in. 


MENSURATION.  29? 

(199)  Compute  the  weight  per  foot  of  an  iron  boiler  tube 
4  inches  outside  diameter  and  3.73  inches  inside  diameter, 
the  weight  of  the  iron  being  taken  at  ,28  pound  per  cubic 
inch.  Ans.   5£  Ib. 

(200)  The  dimensions  of  a  return-tubular  boiler  are  as 
follows:    Diameter,  60  inches;  length  between  heads,  16  feet; 
outside  diameter  of  tubes,  3£  inches;  number  of  tubes,  64; 
distance  of  mean  water-line  from  top  of  boiler,  18  inches. 
(a)  Compute   the  steam  space  of  the  boiler  in  cubic  feet. 
(b}  Determine  the  number  of  gallons  of  water  that  will  be 
required  to  fill  the  boiler  up  to  the  mean  water  level  ? 

Ans    (  (a)  79.Scu.ft 

(  (b)   1,246  gal.,  nearly. 

(201)  The  cylinders  of  a  compound  engine  are   19  and 
31  inches  in  diameter,  and  the  stroke  is  24  inches;  if  the  clear- 
ance at  each  end  of  the  small  cylinder  is  14$  of  the  stroke, 
and  in  the  large   cylinder  8$  of  the  stroke,  (a)  what  is  the 
total  volume  in  cubic  feet  of  the  steam  in  the  small  cylinder 
during  one  stroke  ?     (d)  In   the  large  cylinder  ?     (c )  What 


is  the  ratio  between  the  two  ? 


(a)  4.489  cu.  ft. 

(b)  11.321  cu.  ft. 


Ans. 

'  (c)    Ratio  =  2. 522  :  1. 

(202)  Find  the  volume  of  a  triangular  pyramid  10  inches 
high;  the   base    is   an    equilateral    triangle,  and   one   edge 
measures  10  inches.  Ans.   144.336  cu.  in. 

(203)  The  slant  height  of  a  square  pyramid  is  25  inches, 
and  one  edge  of  its  base  is  16  inches.      Find  its  altitude  by 
the  principle  of  the  right-angled  triangle.     Ans.   23.6854  in. 

(204)  The  length  of  the  circumference  of  the  base  of  a 
cone  is  18.8496  inches,  and  its  slant  height  is  10  inches. 
Find  the  area  of  the  entire  surface  of  the  cone. 

Ans.   122.5224  sq.  in. 

(205)  If  the  altitude  of  the  above  cone  were  9  inches, 
what  would  be  its  volume  ?  Ans.  84.8232  cu.  in. 

(206)  A  square  vat  is  11  feet  deep,  15  feet  square  at  the 
top,  and  12  feet  square  at  the  bottom.     How  many  gallons 
will  it  hold  ?  Ans.   15,058.29  gal. 


298  MENSURATION. 

(207)  How  many  pails  of  water  would  be  required  to  fill 
the  vat,  the  pail  having  the  following  dimensions:     Depth, 
11  inches;  diameter  at  the  top,  12  inches;  diameter  at  the 
bottom,  9  inches?  Ans.  3,627.28. 

(208)  Required  the  area  of  the    convex  surface  of  the 
frustum  of  a  hexagonal  pyramid  whose  slant  height  is  32 
inches,  the  perimeter  of  the  lower  base  measuring  48  inches, 
and  of  the  upper  base  36  inches.  Ans.   1,344  sq.  in. 

(209)  It  is  desired  to  find  the  number  of  gallons  that  a 
certain  tank  will  hold.      By  measuring  with  a  tape  measure, 
it  is  found  to  be  190  inches  in  circumference  at  the  bottom 
and  170^-  inches  in  circumference  at  the  top.     The  depth  of 
the  tank  is  7  feet,  and  the  thickness  of  the  sides  1£  inches. 
Make  the  calculation  in  round  numbers.  Ans.   861  gal. 

NOTE. — To  obtain  the  area  of  the  upper  and  lower  bases,  first  find 
the  outside  diameters  and  then  deduct  the  thickness  of  the  walls,  or 
2  X  li  inches. 

(210)  Find  (a)  the  area  of  the  surface,  and  (b)  the  cubical 
contents  of  a  ball  22£  inches  in  diameter. 

Ans    (  (a)  1,590. 435  sq.  in. 
3'  (  (&]  5,964.1313  cu.  in. 

(211)  What  is  the  volume  of  a  ball  whose  surface  has  an 
area  of  201.0624  square  inches  ?  Ans.   268.0832  cu.  in. 

(212)  (a)  What  is  the  volume  and  area  of  a  cylindrical 
ring  whose  outside  diameter  is  16  inches  and  inside  diameter 
13  inches  ?     (b)  If  made  of  cast  iron,  what  is  its  weight  ? 
Take  the  weight  of  1  cubic  inch  of  cast  iron  as  .261  pound. 

Ans.   Weight  =  21  Ib. 

(213)  The  volume  of  a  certain  cylindrical  ring  is  144.349 
cu.   in.,   its  length  before  bending  (see  line  D  in  Fig.  61, 
Art.  432)  was  20.42  inches;  what  is  the  area  of  its  surface  ? 

Ans.   192.45  sq.  in. 


MECHANICS. 

(SEE  ARTS.  433-558.) 


(214)  (a)  What  is  a  molecule  ?     (b)  An  atom  ? 

(215)  If  a  body  has  an  average  velocity  of  40  feet  per 
second,  how  far  will  it  travel  in  14  minutes  ?  Ans.  6T4T  miles. 

(216)  Show  how  to  represent  a  force  by  a  line. 

(217)  In  Fig.  78,  Art.   491,  let  the  distance  F  c  be  21', 
and  F  b  3£";  what  weight  will  a  force  of  85  Ib.  applied  at 
P  raise  ?  Ans.  510  Ib. 

(218)  What  must  be  the  speed  of  the  driver  pulley  in 
order  that  the  driven  may  make  80  R.  P.  M.   and  be  28*  in 
diameter,  the  diameter  of  the  driver  being  21"  ? 

Ans.  106f  R.  P.  M. 

(219)  The  number  of  teeth  in  a  spur  gear  is  50  and  the 
pitch  is  II" ;  (a)  what  is  the  pitch  diameter  ?  (b)  What  is  the 
outside  diameter?  .         j  (a)  23.87". 

(  (b)  24.77". 

(220)  The  driving  gear  has  45  teeth  and  the  driven  180; 
if  the  driver  makes  212  R.  P.  M.,  how  many  will  the  driven 
make  ?  Ans.  53  R.  P.  M. 

(221)  What  pressure  can  be  exerted  by  a  force  of  24  Ib. 
on  a  half-inch  screw  which  has  13  threads  per  inch,  the  dis- 
tance from  the  center  of  the  screw  to  the  point  on  the 
handle  where  the  force  is  applied  being  11"  ? 

Ans.  21, 563. 94  Ib. 

(#22)  A  ball  weighing  5  Ib.  revolves  in  a  circle  whose 
radius  is  32"  at  the  rate  of  350  R.  P.  M. ;  what  is  the  pull  on 
the  support  caused  by  the  ball  ?  Ans.  555£  Ib. 

(223)  A  body  weighing  2  Ib.  has  a  velocity  of  600  ft.  per 
sec. ;  what  is  its  kinetic  energy  ?  Ans.  11,194  ft.-lb 


300  MECHANICS. 


What  should  be  the  width  of  a  double  leather  belt 
to  transmit  150  horsepower,  when  the  belt  has  a  velocity  of 
3,000  feet  per  minute,  and  has  7  feet  of  its  length  in  contact 
with  the  smaller  pulley,  whose  diameter  is  63"  ?  Give  width 
to  nearest  £'.  Ans.  29.5*. 

(225)  (a)  What  are  the  three  states  of  matter  ?  (/;)  Name 
some  of  the  general  properties  of  matter;    (<:)  some  of  the 
specific  properties. 

(226)  If  a  man  could  run  a  mile  at  the  average  rate  of 
100  yards  in  12  seconds,  how  long  would  it  take  him  ? 

Ans.  3  min.  31.2  sec. 

(227)  What  is  meant  by   "center  of  gravity  "  ? 

(228)  (a)  Why  is  crowning  usually  given  to  the  face  of  a 
pulley  ?     (b)  Why  should  high-speed  pulleys  be  balanced  ? 

(229)  At  what  speed  must  the  engine  run  when  the  di- 
ameter of  the  band-wheel  is  13  feet  and  of  the  main  pulley 
91',  if  the  speed  of  the  main  shaft  is  to  be  108  R.  P.  M.  ? 

Ans.  63  R.  P.  M. 

(230)  The  pitch  of  a  gear  is  24-',  and  the  number  of  teeth 
is  192;  what  is  the  pitch  diameter  ?  Ans.  152.79*. 

(231)  How  many  revolutions  per  minute  must  the  driv- 
ing gear  make  if  it  has  18  teeth,  and  the  driven  has  81  teeth 
and  makes  80  R.  P.  M.  ?  Ans.  360  R.  P.  M. 

(232)  The  nuts  on  a  cylinder  head  are  tightened  by  means 
of  a  wrench  26"  long.     The  threads  in  the  nuts  are  8  to  the 
inch,  and  the  efficiency  of  the  screw  is  40$.     What  pressure 
will  the  nut  exert  against  the  head  when  a  force  of  60  Ib.  is 
applied  to  the  end  of  the  wrench  ?  Ans.  31,365.7  Ib. 

(233)  What  do  you  understand  by  specific  gravity  ? 

(234)  If  the  center  of  gravity  of  a  section  of  an  engine 
fly-wheel  rim  is  6  ft.  If  in.  from  the  center  of  the  shaft,  and 
the  weight  of  the  rim  is  13,000  pounds,   what  is  its  kinetic 
energy  when  making  150  R.  P.  M.?     Ans.  1,883,661.7  ft.  -Ib. 

(235)  What  should  be  the  width  of  a  single  leather  belt 
to  transmit  2^  horsepower  when  the  belt  has  a  velocity  of 


MECHANICS.  301 

2,000  feet  per  minute  ?     The  diameter  of  the  smaller  pulley 
is  14",  and  the  belt  has  18"  of  its  length  in  contact  with  it. 

Ans.  If. 

(236)  (a)  What  is  meant  by  inertia  ?  (/;)  By  weight  ?  (c) 
How  is  weight  measured  ? 

(237)  The  speed  of  a  certain  belt  is  3,000  ft.  per  min. ; 
if  it  drives  a  48"  pulley,  how  long  will  it  take  the  pulley  to 
make  100  revolutions  ?  Ans.  25.13  sec.,  nearly. 

(238)  Find  the  point  of  suspension  of  a  rectangular  cast- 
iron  lever  4  ft.  6  in.  long,  2  in.  deep,  and  £•  in.  thick,  having 
weights  47  and  71  pounds  hung  from  each  end,  in  order  that 
there  may  be  equilibrium.      Take  the  weight  of  a  cubic  inch 
of  cast  iron  as  .261  Ib.  Ang    j  Short  arm  =  22.343*. 

5'  1  Long  arm  =  31.657*. 

(239)  When  two  pulleys  are   used    to    transmit  power, 
which  is  called  the  driven  and  which  the  driver  ? 

(240)  The  driver  is  2  feet  in  diameter  and  the  driven  32" ; 
if  the  driven  makes  63  R.  P.  M.,  how  many  must  the  driver 
make  ?  Ans.  84  R.  P.  M. 

(241)  A  certain  gear  has  a  pitch  of  1£",  and   its  pitch 
diameter  is  11.48";  how  many  teeth  are  in  the  gear  ? 

Ans.  32  teeth. 

(242)  A  fly-ball  governor  is  designed  to  run  at  88  R.  P.  M. 
The  speed  of  the  engine  is  200  R.  P.  M.     The  diameter  of 
the  governor  pulley  is  8" ;  the  number  of  teeth  in  the  bevel 
gear  which  it  'turns,  44,  and  the  number  of  teeth  in  the  other 
bevel  gear,  75;  what  must  be  the  diameter  of  the  pulley  on 
the  crank-shaft  which  drives  the  governor  belt  ?        Ans.  6*. 

(243)  A  bookbinder  has  a  press,  the  screw  of  which  has  4 
threads  to  the  inch.     It  is  worked  by  a  lever  15"  long,  to 
which  is  applied  a  force  of  25  Ib. ;  (a)  what  will  be  the  pres- 
sure  if  the  loss  by  friction  is  5,000  Ib.  ?  .  (b)  What  would  be 
the  theoretical  pressure  ?  A     .    j  (a)  4,424.8  Ib. 

b>  1  (b)  9,424. 8  Ib. 

(244)  A  cubic  foot  of  a  certain  kind  of  wood  weighs  51 
Ib. ;  what  is  its  specific  gravity  ?  Ans.   -816, 


302  MECHANICS. 

(245)  The  piston  of  an  engine  weighs  325  lb.,  including 
the  piston  rod;  what  is  its  kinetic  energy  when  moving  at 
the  rate  of  660  ft.  per  min.  ?  Ans.  611.4  ft.-lb.,  nearly. 

(246)  What  horsepower  can  be  safely  transmitted  by  a 
gear  whose  pitch  is  1",  a  point  on  the  pitch  circle  having  a 
velocity  of  1,200  ft.  per  min.?  Ans.  12  H.  P. 

(247)  (a)  What  is  motion  ?  (b)  Velocity  ?   (c)  Rest  ?    (d) 
Can  a  body  be  in  motion  with  respect  to  one  object  and  at 
rest  with  respect  to  another  ?     Explain  fully. 

(248)  (a)  What    is    force  ?     (b}  Name   several    kinds   of 
forces. 

(249)  Find  by  measurement  the  center  of  gravity  of  a 
triangle  whose  sides  are  4",  5",  and  6"  long. 

Ans.  Ij3/  from  6"  side. 

(250)  The  driving  pulley  makes  40  R.  P.  M. ;  the  driven 
makes  60   R.  P.  M.,  and  is  36"  in  diameter.     What  is  the 
diameter  of  the  driver  ?  Ans.  54". 

(251)  The  diameter  of  the  band-wheel  of  an  engine  is  12 
ft. ;  of  the  main  pulley,  8  ft. ;  of  a  driving  pulley  on  the 
main  shaft,  20";  of  the  driven  pulley  on  the  countershaft, 
6" ;  of  the  driver  on  the  countershaft,  6",  and  of  the  driven 
on   an   emery-wheel  spindle,    4";    if   the   engine   makes  80 
R.  P.  M.,  (a)  what  is  the  speed  of  the  main  shaft  ?     (b)  Of 
the  countershaft  ?     (c]  Of  the  emery  wheel  ? 

(  (a)  120  R.  P.  M. 

Ans.  |  (b)  400  R.  P.  M. 

(  (c)  600  R.  P.  M. 

(252)  The  pitch  diameter  of  a  gear  is  34.15";  if  the  pitch 
is  If",  how  many  teeth  has  the  gear  ?  Ans.  78  teeth. 

(253)  In  order  to  raise  a  weight,  a  combination  of  a  fixed 
and  movable  pulley  is  used;  if  a  force  of  225  lb.  be  applied 
to  the  free  end  of  the  rope,  what  load  will  it  raise  ? 

Ans.  450  lb. 

(254)  If  the  power  moves  through  a  distance  of  5  ft.  6  in. 
while   the  weight  is  moving  6  in.,  (a)  what  is  the  velocity 


MECHANICS.  303 

ratio  of  the  machine  ?     (b)  What  weight  would  a  force  of 
5  Ib.  applied  to  the  power  arm  raise  ?  .         ((«)!!. 

S'  I  (b)  55  Ib. 

(255)  In  the  last  example,   if  the  efficiency   were  65#, 
what  weight  could  the  machine  raise  ?  Ans.  35.75  Ib. 

(256)  How  many  cubic  inches  of  platinum  will  it  take  to 
weigh  10  Ib. ?  Ans.  12.80  cu.  in.,  nearly. 

(257)  (a)  For  what  are  belts  used  ?     (b}  What  is  a  single 
belt  ?     (c)  A  double  belt  ? 

(258)  What  horsepower  can  be  safely  transmitted  by  a 
gear   whose  pitch  is  1.57",  pitch  diameter  30",  and  which 
makes  100  revolutions  per  minute?  Ans.  19.30  H.  P. 

(259)  (a)  What  is  uniform  motion  ?     (b}  What  is  vari- 
able motion  ?     (c)  If  a  body  moves  10  feet  the  first  second, 
12  feet  the  second  second,  15  feet  the  third  second,   etc.,  is 
its  motion  uniform  or  variable,  and  why  ? 

(200)  What  three  conditions  are  required  to  be  known  in 
order  to  compare  forces  ? 

(201)  A  steel  rod  $•*  in  diameter  has  on  one  end  a  cast- 
iron  spherical  ball  5"  in  diameter;  if  the  length  of  the  rod  is 
40"  between  the  ball  and  the  end,  where  is  the  center  of 
gravity  ? 

SUGGESTION. — First  calculate  the  weights  of  the  ball  and  rod  by  the 
aid  of  the  specific  gravity  tables. 

Ans.  0.427"  from  the  center  of  the  ball. 

(202)  The  driving  pulley  makes  240  R.  P.  M.,  and   the 
driven  pulley  180  R.  P.  M. ;  if  the  diameter  of  the  driven 
pulley  is  30",  what  is  the  diameter  of  the  driver  ?     Ans.  22£*. 

(203)  In  a  train  of  gears  used  to  raise  a  weight  of  6,000 
Ib.  in  a  manner  similar  to  that  shown  in  Fig.  83,  Art.  5(K), 
the  diameters  of  the   drivers  and  belt  pulley  are  18",  12*, 
15",  and  12",  and  of  the  pinions  and  drum  6",  5",  8",  and  3*; 
what  force  must  be  applied  to  the  belt  to  raise  the  weight, 
if  20$  of  the  total  force  is  lost  through  friction  ? 

Ans.  138J  Ib. 

(204)  The  pitch  diameter  of  a  gear  is  24.16",  and  the 
number  of  teeth  is  38;  what  is  the  pitch  ?  Ans.  1.997*. 


304  MECHANICS. 

(265)  It  is  required  to  raise  a  load  of  1,890  Ib.  by  means 
of  a  block  and  tackle  which  has  four  fixed  and  four  movable 
pulleys ;  what  force  is  required  to  be  applied  to  the  free  end 
of  the  rope  ?  Ans.  236£  Ib. 

(266)  In  a  block  and  tackle,  the  theoretical  power  neces- 
sary to  raise  a  weight  of  1,000  Ib.  is  50  Ib. ;  (a)  what  is  the 
velocity  ratio  ?     (b)  If  the  actual  power  necessary  to  raise 
the  load  is  95  Ib.,  what  is  the  efficiency  ?    Ang   j  (a)  20. 

'  (  (*)  52.63#. 

(267)  A  piece  of  lead  is  %*  in  diameter  and  10*  long;  how 
much  does  it  weigh  ?  Ans.  12.91  oz. 

(268)  If  the  distance  between  the  centers  of  the  crank- 
shaft and  main  shaft  is  38  ft.,  and  the  diameters  of  the  band- 
wheel  and  main  pulley  are  11  ft.  and  7  ft.,  respectively,  what 
must  be  the  length  of  the  main  belt  ?  Ans.  105  ft.  3*. 

(269)  The  entire  solar  system  is  moving  through  space 
at  the  rate  of  18  miles  per  second;  (a)  what  is  its  velocity  in 
miles  per  hour  ?     (b)  How  far  will  it  go  in  one  day  ? 

Ans   -f  (fl)  64,800  mi.  per  hr. 
5'  I  (b)  1,555,200  mi. 

(270)  (a)  What  is  a  path  of  a  body  ?     (b)  What  line  do 
we  measure  when  we  wish  to  find  the  distance  a  body  has 
traveled. 

(271)  (a)  How  are  forces  measured  ?     (b)  What  kind  of 
an  effect  do  forces  always  tend  to  produce  ? 

(272)  It   is  required  to  raise  a  weight  of  1,500  Ib.   by 
means  of  a  lever  like  that  shown  in  Fig.   71,  Art.  484. 
The  length  of  the  lever  is  4  ft.,  and  the  distance  from  the 
fulcrum  to  the  weight  is  4* ;  what  force  will  it  be  necessary 
to  apply  ?  Ans.  136T4T  Ib. 

(273)  Had  the  lever  in  the  above  example  been  like  that 
shown  in  Fig.  72,  Art.  484,  what  force  would  have  been 
required  ?  Ans.  125  Ib. 

(274)  The  band-wheel  of  an  engine  is  10  feet  in  diameter ; 
what  must  be  the  diameter  of  the  pulley  on  the  main  shaft 
in  order  to  make  110  R.  P.  M.,  if  the  band-wheel  makes  88 
R.  P.  M.?  Ans.  8ft. 


MECHANICS.  305 

(275)  (a)  What  is  a  spur  gear  ?     (b)  A  miter  gear  ?     (c) 
A  bevel  gear  ? 

(276)  The  pitch  diameter  of  a  gear  is  30.56",  and  the 
number  of  teeth  is  42;  what  is  the  pitch  ?  Ans.   2.735*. 

(277)  The  length  of  an  inclined  plane  is  400  feet,  and  the 
height  is  45  feet;  what  force  acting  parallel  to  the  plane  will 
be  required  to  pull  up  the  plane  a  weight  of  4,000  Ib.  ? 

Ans.   450  Ib. 

(278)  (a)  What    is    meant    by    the    velocity    ratio    of   a 
machine  ?    (b}  By  the  efficiency  ? 

(279)  If  a  coil  of  brass  wire  weighs  10  Ib. ,  and  the  diam- 
eter of  the  wire  is  Ty,  how  Long  is  the  wire  ? 

Ans.   806  ft.,  nearly. 

(280)  The  diameters  of  two  pulleys  are  14*  and  18*,  and 
the  distance  between  their  centers  is  14  feet;  what  must  be 
the  length  of  a  belt  to  drive  these  pulleys  ?     Ans.   32  ft.  4". 

(281)  A  velocity  of  30  miles  per  hour  corresponds  to  how 
many  feet  per  second  ?  Ans.  44  ft.  per  sec. 

(282)  The  stroke  of  a   steam  engine  is  28*,  and  it  makes 
1,500  strokes  in  6  minutes;  what  is  the  velocity  of  the  piston 
in  feet  per  second  ?  Ans.  Off  ft.  per  sec. 

(283)  State  the  three  relations  between  force  and  motion. 

(284)  A  pulley  on  the  main  shaft  is  40*  in  diameter,  and 
makes  120  R.  P.  M. ;  what  must  be  the  diameter  of  a  pulley 
on  the  countershaft  that  is  to  make  160  R.  P.  M.  ? 

Ans.   30'. 

(285)  (a)  What  is  a  rack  ?  (b)  A  worm-wheel  ?  (c)  A  worm  ? 

(286)  (a)  What  distinguishes  the  epicycloidal  teeth  from 
the  involute  teeth  ?  (b}  Name  two  advantages  which  the 
latter  possess  over  the  former. 

(287)  An  inclined  plane  has  a  length  of  1,200  feet  and  a 
height  of  125  feet.      It  is  required  to  pull  a  load  of  50,000  Ib. 
up    this    plane.       A    block   and  tackle  having  6  fixed  and 
6  movable  pulleys  is  stationed  at  the  top  of  the  plane,  and  the 
weight  end  of  the  rope  is  attached  to  the  load.      If  the  rope 
which  connects  the  block  to  the  load  is  parallel  to  the  plane. 


306  MECHANICS. 

what  force  will  it  be  necessary  to  exert  on  the  free  end  of  the 
rope  to  pull  up  the  load,  no  allowance  being  made  for  friction? 

Ans.   434  Ib. 

(288)  (a)  What  do  you  understand  by  centrifugal  force  ? 
(b)  By  centripetal  force  ? 

(289)  Define    (a)    work;    (b)     horsepower;     (c)    kinetic 
energy ;  (d)  potential  energy. 

(290)  Two  pulleys  have  diameters  of  8"  and  20";  the  dis- 
tance between  their  centers  being  19  ft.  3",  what  must  be  the 
length  of  a  belt  to  drive  them  ?  Ans.   42  ft.  3£*. 

(291)  Two  bodies  starting  from  the  same  point,  move  in 
opposite  directions,  one  at  the  rate  of  11  feet  per  second,  and 
the  other,  15  miles  per  hour;  (a)  what  will  be  the  distance 
between  them  at  the  end  of  8  minutes;  (b)  How  long  before 
they  will  be  825  feet  apart  ?  Ang   j  (a)  3  miles. 

'  1  (b)  25  seconds. 

(292)  A  railroad  train  runs  2  miles  in  2  minutes  and  10 
seconds ;  what  is  its  average  velocity  in  feet  per  second  ? 

Ans.   81.23  ft.  per  sec. 

(293)  Why  is  it  difficult  to  jump  from  a  rowboat  into  the 
water  ? 

(294)  A  compound   lever,    similar  to  the  one  shown  in 
Fig.  77,  Art.  49O,  is  required  to  lift  a  weight  of  1,250  Ib. 
The  lengths  of  the  power  arms  P  F  are  30*,  20",  10",  and  15", 
of  the  weight  arms  W  F  6",  5",  4",  and  7";  what  force  will  be 
required?  Ans.   llf  Ib. 

(295)  The  driving  pulley  is  20"   in  diameter  and  driven 
16";  if  the  driver  makes  150  R.    P.  M.,   how  many  will  the 
driven  make  ?  Ans.   187£  R.  P.  M. 

(296)  How  is  the  diameter  of  a  gear  measured  ? 

(297)  The  driving  gear  makes   100  and  the   driven,  40 
R.  P.  M. ;    if  the  driven  has  60  teeth,   how  many  has  the 
driver  ?  Ans.  24. 

(298)  The  base  of  an  inclined  plane  is  80  feet  long,  and 
the  height  of  the  plane  is  50  feet;  what  force  exerted  parallel 
to  the  base  will  raise  a  load  of  750  Ib.  ?  Ans.  468|  Ib. 


MECHANICS.  307 

(299)  What  is  the  centrifugal  force  of  the  counterweight 
of  a  steam  engine,  the  counterweight  weighing  128  Ib. ,  and 
its  center  of  gravity  being  8f  from  the  center  of  the  shaft  ? 
The  crank  makes  180  R.  P.  M.  Ans.    1,028.16  Ib. 

(300)  How  much  work  can  be  done  by  20  cubic  feet  of 
water  falling  from  a  height  of  50  feet  ?      Ans.   G2,500  ft.-lb. 

(301)  What  horsepower  can  be  transmitted  by  a  single 
leather  belt  5"  wide,  which  runs  at  the  rate  of  1,960  ft.  per 
min.  ?     The  diameter  of  the  smaller  pulley  is  15"  and  the 
length  of  the  arc  of  contact  is  21"  ?     Ans.   11.4  H.  P.,  nearly. 

(302)  It  is  required  to  raise  a  weight  of  18,000  Ib.  by  means 
of  a  screw  having  3  threads  per  inch ;  if  the  length  of  the 
handle  is  15",  and  there  is  a  loss  of  10,000  Ib.  due  to  friction, 
etc.,  what  force  will  it  be  necessary  to  apply  to  the  handle  ? 

Ans.   99  Ib. ,  nearly. 

(303)  The  fly-wheel  of  an  engine  is  9  feet  in  diameter 
(outside);  if  the  fly-wheel  makes  100  R.    P.  M.,  how  many 
miles  will  a  point  on  the  rim  travel  in  1^  hours  ? 

Ans.   40.16£  miles. 

(304)  Suppose  that  an  air  gun  can  throw  a  ball  with  a 
velocity  of  100  feet  per  second,  and  that  a  man  standing  on 
a  railroad  train,  which  is  moving  at  the  rate  of  100  feet  per 
second,  were  to  fire  the  gun  in  a  direction  exactly  opposite 
to  that  in  which  the  train  is  moving,  what  would  become  of 
the  ball  ?     Why  ? 

(305)  If  the  distance  between  the  center  line  of  the  handle 
and  the  axis  of  the  drum  shown  in  Fig.  79,  Art.  49 1 ,  is  1 4£", 
and  the  diameter  of  the  drum  is  5",  what  load  will  a  force  of 
30  Ib.  exerted  on  the  handle  raise  ?  Ans.   174  Ib. 

(306)  A  pulley  on  the  main  shaft  is  42'  in  diameter,  and 
makes  108  R.  P.  M. ;  what  will  be  the  speed  of  the  counter- 
shaft if  the  driven  pulley  is  36"  in  diameter  ? 

Ans.   126  R.  P.  M. 

(307)  What  is  (a)  the  pitch  circle  ?  (If)  The  pitch  of  a 
gear? 


308  MECHANICS. 

(308)  The  driving  gear  makes  360  R.   P.   M.,  and  the 
driven  170  R.  P.  M. ;  if  the  driver  has  34  teeth,  how  many 
has  the  driven  ?  Ans.   72  teeth. 

(309)  Name  some  particular  use  in  the  engine  room  or 
shop  to  which  you  have  seen  the  inclined  plane  put. 

(310)  The  mean  diameter  of  the  rim  of  an  engine  cast- 
iron  fly-wheel  is  9  ft.  lOf,  its  width  is  22",  and  its  thickness 
2£";    what  is  the  centrifugal    force  when   running  at    210 
R.  P.  M.?  Ans.  397,450  Ib. 

(311)  Assuming  the  average  pressure  upon  the  piston  of 
a  steam  engine  to  be  41.38  pounds  per  square  inch,  what  is 
the  horsepower  ?     The  diameter  of  the  piston  is  10* ;  the 
stroke  16",  and  the  number  of  strokes  per  minute  450. 

Ans.  59.091  H.  P.,  nearly. 

(312)  What   horsepower  can   be    transmitted    by   a   20* 
double  leather  belt  which  has  a  velocity  of  2,800  ft.  permin.  ? 
The  diameter  of  the  smaller  pulley  is  4  ft.,  and  the  belt  has 
5  ft.  9  in.  of  its  length  in  contact  with  it.      Ans.  99.4  H.  P. 


MECHANICS. 

(SEE  ARTS.  559-655.) 


(313)  State  Pascal's  law. 

(314)  A  cylinder  fitted  with  a  piston  is  used  as  a  lifting 
cylinder  by  passing  a  rope  over  a  pulley  and  fastening  one 
end  to  the  piston  rod.     The  piston  is  moved  by  means  of 
water  obtained  from  the  city  reservoir,  and  a  gauge  attached 
to  a  pipe  near  the  cylinder  shows  the  pressure  to  be  90  Ib. 
per  sq.  in.     The  diameter  of  the  cylinder  is  19  in.  and  of  the 
pipe  £  in.      If  friction  be  neglected,  how  great  a  weight  can 
be  raised  ?  Ans.  25,517.0  Ib. 

(315)  Give  Mariotte's  law. 

(31G)     What  do  you  understand  by  (a)  the  tensile  strength 
of  a  material  ?  (£)  The  working  stress  ? 

(31?)     A  close-link  wrought-iron  chain  is  made  from  f* 
iron;  what  is  the  greatest  safe  load  that  it  will  carry  f 

Ans.   1,087.5  Ib. 

(318)  What  is  the  allowable  working  load  for  a  steel-wire 
rope  5i*  in  circumference  ?  Ans.  27,502.5  Ib. 

(319)  What  steady  force  is  required  to  shear  a  steel  crank- 
pin  which  is  6"  in  diameter  ?  Ans.  1,090,404  Ib. 

(320)  What  must  be  the  diameter  of  a  wrought-iron  shaft 
to  transmit  40  horsepower  when  running  at  110  revolutions 
per  minute,  no  power  being  taken  off    between  bearings. 
Ans.  2.89'.     A  shaft  2}|*  in  diameter  would  be  used,  that  is, 
3*  round  iron  turned  down. 

(321)  In  Fig.   104,  Art.  562,  suppose  that  the  area  of 
the  piston  c  is  .G  sq.  in.  ;  of  d,  3  sq.  in. ;  of  e,  0  sq.  in.  ;  of/, 
2  sq.  in. ;  of  a,  14  sq.  in.,  and  of  b,  9  sq.  in. ;  if  a  force  of 


310  MECHANICS. 

24  lb.  be  applied  to  c,  what  must  be  the  forces  applied  to  the 
other  pistons  to  counterbalance  the  force  at  c,  neglecting 
the  weight  of  the  water  and  the  weight  of  the  pistons  ? 

f  At  d,  120  Ib. 

At  e,  240  Ib. 

Ans.  J  At/,    80  Ib. 

At  a,  5GO  Ib. 

[  At  b,  360  Ib. 

(322)  The  upper  base  of  a  cylinder  submerged  in  water 
is  40  ft.  below  the  surface.      The  diameter  of  the  cylinder 
is  20  inches,  the  altitude  36  inches,  and  the  bases  are  parallel. 
If  the  bases  are  horizontal,  what  is  (a)  the  upward  pressure 
of  the  water  on  the  cylinder  ?    (b)  The  downward  pressure  ? 

Ans   \  <*)  5'863'39  lb' 
I  (b)  5,454.32  lb. 

(323)  (a)     What    is    a    barometer  ?     (b}     What   is   the 
essential    difference   between    a    mercurial  and   an  aneroid 
barometer  ? 

(324)  The  smallest  section  of  a  connecting-rod  is  3.5  sq. 
in. ;  what  is  the  unit  stress  when  subjected  to  a  tensile  stress 
of  12,400  lb. ?  Ans.   3,543  lb.  per  sq.  in.,  nearly. 

(325)  What  load  may  be  safely  carried  by  a  hemp  rope 
4*  in  circumference  ?  Ans.   1,600  lb. 

(326)  What   load  can    be  safely   sustained   by   a  round 
wooden  pillar,  8"  in  diameter  and  10  ft.  long,  having  both 
ends  flat  ?  Ans.  13£  tons. 

(327)  What  force  is  required  to  punch  a  I"  hole  through 
a  wrought-iron  plate  T7T*  thick  ?  Ans.   54,978  lb. 

(328)  What    horsepower  will   a   1£*   wrought-iron    shaft 
transmit  when  running  at   180  revolutions  per  minute,  an 
average    amount    of   power   being   taken    off    between    the 
bearings  ?  Ans.  12.49  H.  P. 

(329)  Does  the  shape  of  the   vessel  have  any  effect  in 
regard  to  the  pressure  exerted  by  a  liquid  upon  its  bottom  ? 

(330)  In  Fig.  Ill,  Art.  576,  suppose  that  the  diameter 
of  the  piston  a  is  2",  and  of  b  7$';  if  a  weight  of  400  lb.  be 


MECHANICS.  311 

laid  upon  b,  what  weight  must  be  applied  to  a  to  balance  the 
weight  on  b  ?  Ans.  28.44  +  Ib. 

(331)  A  vessel  contains  42  cu.  ft.  of  coal  gas  having  a 
tension  of  20£  Ib.  per  sq.  in. ;  what  will  be  the  new  tension 
when  allowed  to  communicate  with  a  perfectly  empty  vessel 
whose  volume  is  14  cu.  ft.?  Ans.  15.19  Ib.  per  sq.  in.,  nearly. 

(332)  What  safe  steady  load  can  be  sustained  by  a  1£* 
round  wrought-iron  bar,  the  load  producing  a  tensile  stress  ? 

Ans.   21,205.2  Ib. 

(333)  What  load  can  a  hemp  rope  0"  in  circumference 
carry  with  safety  ?  Ans.   3,600  Ib. 

(334)  What  load   will   a  hollow  cast-iron  pillar  support 
with  safety  if  the  pillar  is  20  ft.  long,   outside  diameter  14", 
inside  diameter  11^-*,  and  both  ends  are  fixed  ? 

Ans.   219.24  tons. 

(335)  What   force   is   required  to   punch   a  hole  \\"  in 
diameter  through  a  f  steel  plate  ?  Ans.   212,058  Ib. 

(330)  A  vessel  is  filled  with  water  to  the  depth  of  18"; 
if  the  area  of  the  bottom  is  46  sq.  in.,  what  is  the  total 
pressure  on  the  bottom  ?  Ans.  29.95  Ib. 

(337)  Why  does  water  seek  its  level  ? 

(338)  The   volume  of  steam  in   an    engine    cylinder    is 
feu.  ft.  at  cut-off,  and  its  pressure  is   94.7  Ib.  per  sq.  in.; 
what  will  be  its  pressure  when  the  volume  has  increased  to 
2£  cu.  ft.?  Ans.  23.675  Ib.  per  sq.  in. 

(339)  A  bar  of  steel  having  a  cross-section  of  If*  X  3'  is 
subjected    to   a    tensile    stress;  if   the    stress    is    suddenly 
applied,  what  is  the  greatest  load  that  it  will  safely  carry  ? 

Ans.   31,500  Ib. 

(340)  A  load  of  2,400  Ib.  is  to  be  raised  by  means  of  a 
hemp  rope;  what  should  be  the  circumference  of  the  rope  ? 

Ans.   4.9'. 

(341)  Regarding  the  piston  rod  of  a  steam  engine  as  a 
pillar  which  has  one  end  flat  and  the  other  round,  what 
should  be  the  greatest  diameter  of  the  piston  if  the  rod  is 
4  ft.  8  in.  long,    3£  in.  in  diameter,  and  the  greatest  steam 


312  MECHANICS. 

pressure  is  not  to  exceed  100  Ib.  per  sq.  in.  ?     The  rod  is 
made  of  wrought  iron.  Ans.  25*,  nearly. 

(342)  (a)   What    is    cold-rolled     shafting?      (b)    Bright 
shafting  ?     (c)  Black  shafting  ? 

(343)  A   tank   contains   cylinder    oil    having   a   specific 
gravity  of  say  .92.     This  tank  is  connected  to  a  four-gallon 
can  by  means  of  a  pipe  whose  internal  diameter  is  f .     If 
the  vertical  distance  between  the  level  of  the  oil  in  the  tank 
and  the  end  of  the  pipe  is  8  ft.  7  in.,  what  is  the  pressure 
per  square  inch  at  the  end  of  the  pipe  ? 

Ans.   3.427  Ib.  per  sq.  in. 

(344)  In   Fig.    113  (see  examples  following  Art.  578), 
suppose  that  the  diameter  of  the  plunger  A  is  12",  and  of 
the  plunger  C  of  the  pump  B  %" ;  if  a  force  of  100  Ib.  be 
applied  to  C  by  means  of  the  lever  D,  how  great  a  weight 
can  the  plunger  A  raise  if  the  weight  of  A  is  600  Ib.  ? 

Ans.   58, 382. 4  Ib. 

(345)  The  volume  of  steam  in  an  engine  cylinder  at  the 
beginning  of  compression  is   1.11  cu.  ft.,  and  its  pressure  is 
18  Ib.  per  sq.  in.     At   the  end  of  compression  the  volume 
is  .3  cu.  ft. ;  what  is  the  final  pressure  ? 

Ans.   66.6  Ib.  per  sq.  in. 

(346)  What  should  be  the  least  area  of  one  of  the  14 
wrought-iron   cylinder   head  stud  bolts  if  the  diameter  of 
the  cylinder  is  19',  and  the  greatest  steam  pressure  is  180  Ib. 
per  sq.  in.  ?     Assume  that  the  studs  are  subjected  to  shocks. 

Ans.   .729  sq.  in. 

(347)  What  should  be  the  circumference  of  a  hemp  rope 
to  safely  sustain  a  load  of  4,200  Ib.?  Ans.   G£". 

(348)  Regarding  the  connecting-rod  of  a  steam  engine 
as  a  pillar  with  two  round  ends,  what  is  the  greatest  force 
that  may  be  exerted  on  the  cross-head  if  the  connecting-rod 
is  made  of  wrought  iron,  is  10  ft.  long,  and  has  a  rectangular 
cross-section  6*  by  Z±*  ?  Ans.   35,489  Ib. 

(349)  If    you    were    to    order    some    2£*   bright-turned 
shafting,  what  size  would  you  expect  to  get  ? 


MECHANICS.  313 

(350)  A  tank  having  the  shape  of  a  frustum  of  a  cone  is 
filled  with  water.     If  the  diameter  of  the  large  end  is  8  ft., 
of  the  small  end  6  ft.,  and  the  perpendicular  distance  be- 
tween the  two  ends  is  10",  (a)  what  is  the  pressure  on  the 
bottom  when  the  large  end  is  down  ?     (b]     When  the  small 
end  is  down  ?  .         (  (a)  2,618  Ib. 

S'  (  (If)  1,473  Ib.,  nearly. 

(351)  What  is  the  total  pressure  on  all  of  the  six  sides  of 
a  cube  which  has  been  sunk  in  the  water  until  its  top  is 
50  ft.  below  the  surface  ?     One  edge  of  the  cube  measures 
2  ft.,  and  its  upper  base  is  parallel  with  the  water  level. 

Ans.  70,502  Ib. 

(352)  A  vessel  contains  25  cu.  ft.  of  air  having  a  pres- 
sure of  45  Ib.   per  sq.   in.     When  allowed  to  communicate 
with  a  second  vessel  which  is  entirely  empty,  the  pressure 
falls  to  15  Ib.  per  sq.  in.     What  is  the  volume  of  the  second 
vessel  ?  Ans.  50  cu.  ft. 

(353)  A  steam  cylinder  is  44"  in  diameter  and  sustains  a 
steam  pressure  of  100  Ib.  per  sq.  in.     The  diameter  of  the 
cylinder  head  studs  is  If  and  the  area  at  the  bottom  of 
the  thread  is  1.057  sq.  in.     How  many  wrought-iron  studs 
are  required  ?     Assume   that   the   studs   are   subjected   to 
shocks.  Ans.  29  studs. 

(354)  An  iron-wire  rope  4*  in  circumference  is  used  on  a 
crane  for  hoisting  loads;  what  is  the  greatest  load  that  the 
rope  will  sustain  with  safety  ?  Ans.  9,  GOO  Ib. 

(355)  A  cast-iron  rectangular  cantilever  beam  having  a 
cross-section  of  1£*  wide  by  2£"  deep  is  4  ft.  8*  long;  how 
great  a  weight  will  the  beam  sustain  at  its  end  ? 

Ans.  201  Ib.,  nearly. 

(350)  What  horsepower  will  a  2TV  steel  shaft  transmit, 
when  running  at  120  revolutions  per  minute,  pulleys  being 
carried  between  the  bearings  to  distribute  the  power  along 
the  line  ?  Ans.  20.445  H.  P. 

,  ,(357)     A  board  8*  X  20*  and  If  thick  is  so  placed  in  water 
that  its  flat  sides  are  horizontal;  if  the  distance  from  the 


314  MECHANICS. 

level  of  the  water  to  the  top  of  the  board  is  5  feet,  what  is 
the  total  upward  pressure  on  the  board  ?     Ans.  355.91  -f-  Ib. 

(358)  Will  any  solid  body  whose  specific  gravity  is  greater 
than  that  of  water,  sink  in  it  to  any  depth  ?     Why  ? 

(359)  The  volume  of  steam  in  an  engine  cylinder  at  cut- 
off is  1.6  cu.  ft.,  and  its  pressure  is  90  Ib.  per  sq.  in. ;  what 
must  be  its  volume  at  release  when  the  pressure  has  fallen 
to  21  Ib.?  Ans.  Gf  cu.  ft 

(360)  What  should  be  the  least  diameter  of  a  wrought- 
iron  bolt  that  is  to  resist  a  sudden  pull  of  12,000  Ib.  ? 

Ans.  1.74"+. 

(361)  A  steel-wire   rope    is   41"  in  circumference ;  what 
load  will  it  safely  sustain  ?  Ans.  22,562.5  Ib. 

(362)  A  white-pine  beam  supported  at  both  ends  has  a 
rectangular  cross-section  8"  wide  by  10"  deep;  if  the  beam 
is  28  ft.   long,   what  total  uniform  load  will  it  support  in 
safety  ?  Ans.  6,857|  Ib. 

(363)  What  horsepower  can  a  10"  wrought-iron  crank- 
shaft transmit,  when  running  at  200  revolutions  per  minute  ? 

Ans.  2,857|. 

(364)  A  vertical  cylinder  having  a  diameter  of  20"  and  a 
length  inside  of  36"  is  filled  with  water  ;    a  pipe  having  a 
diameter  of  |"  is  screwed  into  the  upper  head  and  fitted  with 
a  piston  weighing  10  oz.,  on  which  is  laid  a  weight  of  25  Ib.  ; 
if  the  end  of  the  pipe  is  10  ft.  above  the  level  of  the  water 
in  the  cylinder,  (a)  what  is  the  pressure  per  square  inch  on 
the  top  of  the   cylinder  ?     (b]  On  the  bottom  ?     (c)  What 
equivalent  weight  laid  on  the    lower  cylinder  head  would 
replace  the  pressure  it  sustains?     ,  (^)  330.45  Ib.  per  sq.  in. 

Ans.  |  (£)  237.75  Ib.  per  sq.  in. 
(   (c)  74,691. 54  Ib. 

(365)  If,  in  the  last  example,  a  hole  one  inch  in  diameter 
is  drilled  through  the  cylinder  wall  midway  of  its  length, 
and  covered  by  a  flat  plate  in  such  a  manner  that  the  water 
can  not  leak  out,  what  would  be  the  pressure  against  the 
plate  ?  Ans.  186.22  Ib. 


MECHANICS.  315 

(3GG)     The  vacuum  gauge  of  a  condensing  engine  indi- 
cates 20"  of  vacuum ;  what  is  the  pressure  in  the  condenser  ? 

Ans.  4.9  Ib.  per  sq.  in. 

(367)  The  volume  of  the  receiver  of  an  air  compressor 
is  300  cu.  ft.,  and  the  pressure  of  the  air  which  is  contained 
in  it  is  52  Ib.  per  sq.  in. ;  if  120  cu.  ft.  of  air,  having  a  pres- 
sure of  52  Ib.  per  sq.   in.,  are  removed  from  the  receiver, 
what  will  be  the  pressure  of  the  air  which  remains  ? 

Ans.  31.2  Ib.  per  sq.  in. 

(368)  What  is  the  greatest  safe  load  that  may  be  applied 
to  a  stud-link  wrought-iron  chain  if  the  diameter  of  the  iron 
from  which  the  link  is  made  is  £"  ?  Ans.  4,500  Ib. 

(369)  It  is  desired  to  handle  loads  up  to  14,000  Ib.  by 
means  of  an  iron-wire  rope;    what  should  be  its    circum- 
ference ?  Ans.  4.83",  nearly. 

(370)  What  is  the  greatest  load  that  a  bar  of  wrought 
iron  2"  in  diameter  and  6  ft.  long  can  safely  sustain  in  the 
middle  ?     The  bar  is  merely  supported  at  its  ends. 

Ans.  480  Ib. 

(371)  What  must  be  the  diameter  of  a  cast-iron  crank- 
shaft to  transmit  1.000  horsepower  at  80  revolutions  per 
minute  ?  Ans.  10.4  in. 

(372)  (a)  What    must   be    the    height    of    the    mercury 
column  to  indicate  a  vacuum  of  12"?     (b)  Of  18*? 

(373)  (a)  What  is  a  stress?  (b)  A  strain?  (c)  A  unit  stress? 

(374)  The  links  in    a  stud-link  wrought-iron  chain  are 
made  from  iron  |f  m  diameter;  what  is  the  greatest  safe 
load  that  the  chain  can  handle?  Ans.   11,883  Ib. 

(375)  A  steel-wire  rope  is  used  to  haul  loads  up  an  inclined 
plane ;  the  greatest  stress  in  the  rope  is  8,000  Ib. ;  what  should 
be  its  circumference?  Ans.  2.83*. 

(376)  What  uniform  load  can  be  safely  sustained  by  a 
steel  beam  20  ft.  long,  2"  wide,  and  6"  deep?    Ans.  4,608  Ib. 

(377)  A  4"  steel  shaft  is  to  transmit  80  horsepower;  how 
many  revolutions  per  minute  must  it  make  if  used  for  trans- 
mission only  (that  is,  no  power  being  taken  off  at  intermediate 
points)?  Ans.  81±  R.  P.  M. 


316  MECHANICS. 

(378)  Define  tension  as  applied  to  gases. 

(379)  (a)  What  is  elasticity?  (b)  Elastic  limit?  (c)  What 
is  meant  by  set? 

(380)  What  safe  load   may  be   carried    by    a    close-link 
wrought-iron  chain  whose  links  are  made  from  f  iron? 

Ans.  4,687.5  Ib. 

(381)  What  is  the  allowable  working  load  for  an  iron-wire 
rope  6"  in  circumference?  Ans.  21,600  Ib. 

(382)  What  force  is  required  to  shear  a  wrought-iron  strip 
4  ft.  long  and  £"  thick?  Ans.  960,000  Ib. 

(383)  A  7"  wrought-iron  crank-shaft  is  to  transmit  200 
horsepower;  how  many  revolutions  per  minute  must  it  make? 

Ans.  40.8  rev.,  nearly. 


INDEX. 


A.                                  PAGE 

PAGE 

Abstract  number        .... 

i 

Area  of  frustum  of  pyramid     . 

M4 

Acute  angle         

120 

"     "  irregular  polygon 

I3« 

Addition      

4 

"     "  parallelogram 

-134 

"        of  decimals. 

40 

"     "  prism     .        .        .        . 

:     » 

"         "  denominate  numbers   . 

68 

"     "  pyramid         .... 

143 

"         "  fractions 

29 

"     "  regular  polygon   . 

»3« 

"•         Proof  of       .... 

8 

"     "  sector  of  circle      . 

••  n 

"         Rule  for 

8 

"     "  segment  of  circle 

137 

Sign  of        .... 

4 

"     "  trapezoid        .... 

104 

table    

5 

"     "  triangle          .... 

tag 

Adjacent  angle  .        .        .        .        . 

120 

Atmosphere,  Pressure  of  . 

223 

Aeriform  bodies         .... 

'5° 

Atoms  

>49 

Aggregation,  Symbols  of 

53 

Avoirdupois  weight  .... 

63 

Air  chamber       

244 

"  compressors         .... 

235 

B.                                  PAGE 

"  pump     

231 

Balanced  pulleys       .... 

'-> 

Altitude  of  cone          .... 

142 

Barometer  

MM 

"          "  frustum  of  cone  or  cyl- 

"          Aneroid    .... 

tat 

inder      .... 

144 

"           Mercurial 

waj 

"          "    parallelogram 

123 

Base  (in  percentage)  .... 

yi 

"          "    pyramid 

142 

"     of  parallelogram 

123 

"          "   triangle  .... 

129 

"      "  plane  figure  .... 

133 

Amount  (in  percentage)    . 

56 

Beams,  Transverse  strength  of 

rfa 

Aneroid  barometer    .... 

225 

Bearings  for  line  shafting,  Distance 

Angle   

120 

apart  of    

•"•. 

"     Acute         

120 

Belts     

201 

"     Adjacent  

120 

"      Double       

.•    i 

"     Inscribed  

133 

"      Horsepower  of          ... 

t  * 

"     Measured  by  arc      . 

121 

"      Lacing        

1  .; 

"     Obtuse       

120 

"      Length  of  

.-    I 

"     Right          

120 

"      Rules  for            .... 

"     Vertex  of  

120 

"      Single         

|  I] 

Angles  or  arcs.  Measures  of     . 

64 

"      Width  of    

Antecedent  (of  a  ratio) 

97 

Bevel  gears         

T- 

Arabic  notation  

2 

Black  shafting    

.•- 

Arc  of  circle       ..... 

'33 

Block  (pulley)     

I83 

"    "      "       To  find  length  of 

136 

Bodies,  Aeriform        .... 

'      • 

Archimedes,  Principle  of  . 

220 

"       Gaseous         .... 

..  , 

Area  of  a  surface       .... 

124 

how  composed 

[90 

"     "  circle      

136 

Liquid    

'  -  • 

"     "  circular  ring 

I36 

Solid      

IV 

"     "  cone       

142 

Brace    53, 

116 

"     "  cylinder        .... 

'39 

Bracket        53, 

116 

"     "  cylindrical  ring    . 

146 

Bright  shafting  

u      "  frustum  of  cone    . 

144 

Brittleness  . 

i     ; 

Buoyant  effect  of  water  . 
Bushel,  Cubic  inches  in  . 
Butt  joint  .  .  . 


Cancelation 

"          Rule  for 
Cantilever 

"          Strength  of 
Capacity,  Measures  of 
Cause  and  effect         . 
Center  of  gravity       . 


PAGE 

.219 
.65 

.316 

PAGE 
21 

.      23 
261 

.  263 
.64 
.  108 
.  160 


"      "        "        of  irregular  plane 

figure     .        .     162 
"      "       "         "  parallelogram.     162 
"      "        "         "  regular  p  1  a  n  e 

figure     .        .     162 
"      "        "         "  solid  .        .        .164 
"      "        "         "  system  of  bod- 

ies .        .        .     161 

"      "        "         "  triangle     .        .     162 
Centrifugal  force        ....     194 

"  "     Rule  for         .        .     195 

Centripetal  force        ....     195 

Chains  .......    251 

"       Rules  for  strength  of    .        .    251 

"       Treatment  of         ...    251 

Chamber,  Air      .....    244 

Chord  of  circle   .....    133 

Cipher          .....  • 

Circle  ......      120,  133 

"      Arc  of       ....      121,  133 

"      Area  of    .....     136 

"      Center  of          ...      120,  133 
"      Chord  of  .        .        .        .        .     133 

"      Circumference  of   .        .      120,  133 
"      Diameter  of    .        .        .        .     133 

"      Divisions  of    .        .        .        .121 

"      Radius  of         .  .      .        .        .    133 

"      Sector  of  .....     136 

"      Segment  of      .        .        .        .137 

Circular  ring.  Area  of       .        .       '.     136 
Circumference  of  circle    .        .        .     133 
Coefficient  of  friction         .        .        .     189 
Table  of        .     192 
Cold-rolled  shafting  .        .        .        .268 

Combination  of  pulleys    .        .        .     184 

"  "        "        Law  of        .     185 

Common  denominator       ...      27 

"  "  Least     .        .      27 

Composite  numbers  ....      21 

Compound  denominate  numbers    .      62 

lever        .        .  .167 

"  proportion       .        .        .     109 

Compressibility          .        .        .        .152 

Compressive  strength  of  materials.    255 


Compressor,  Air 

Concrete  number 

Cone 

"      Altitude  of 
"      Convex  area  of 
"      Frustum  of 
"      Slant,  Height  of 
"      Vertex  of 
"      Volume  of 

Consequent 

Conservation  of  energy    . 


PAGE 

•  235 

.  142 

.  142 

.  142 

•  143 
.  142 
.  142 
.  142 

•  97 

.  200 

Constants  for  cast-iron  pillars          .  258 

"          "    line  shafting       .        .  270 

"          "    transverse  strength  .  261 

"           "    wooden  pillars   .        .  259 

"    wrought-iron  pillars .  257 

Convex  area  of  cone  ....  142 

"          "     "  cylinder    .        .        .  139 

"          "     "  cylindrical  ring       .  146 

"          "     "  frustum    .        .        .  144 

"    "  prism        .        .        .139 

"     "  pyramid   .        .        .  142 

"           "     "  sphere       .        .        .  145 

"           "     "  surface  of  solid        .  139 

Countershafts 268 

Couplet  (ratio  and  proportion)        97,  100 

Cross-head  friction  in  guides  .        .  193 

Crowning  of  pulleys          .        .        .  171 

Crushing  strength  of  materials       .  255 

"Table  of  256 

Cube  (solid) 138 

"     of  a  number      .        .               ..  77 

"     root 79,  86 

"       "    Proof  of      ....  92 

"        "    Rule  for    .        ...  90 

Cubic  measure 63 

Cubical  contents         ....  139 

Curved  line 119 

Cylinder,  Convex  surface  of    .        .  139 

"               "        Volume  "    .        .  139 

Cylindrical  ring,  Convex  area  of    .  146 

"            "      Volume  of    .        .  147 

D.  PAGE 

Decimal 38 

"       number,  how  read      .        .  38 
"       part  of  a  foot.  To  reduce 

inches  to  .        .        .        .50 

"       point 38 

"  To  express  as  a  fraction 
with  a  given  denomi- 
nator .  .  .  .52 

"       To  reduce  a  fraction  to    .  50 

"         "        "       to  a  fraction     -  51 

Decimals,  Addition  of               .        .  40 
"          Division  of                .        -45 


PAGE 

Decimals,  Multiplication  of      .        .43 
"          Subtraction  of  .        .        .42 

PAGE 

Expansibility      152 
Exponent    77 

"        and  limits  of  exhaustion  . 

233 

"          Measures  of      ... 

o* 
6» 

Denominate  numbers 

61 

Extremes  of  a  proportion 

tat 

"                 "         Addition  of   . 

68 

"                  "         Compound     . 

62 

F.                                 PAGE 

"                  "         Division  of     . 

73 

Factor  

-.-I 

"                  "          Multiplication 

"       Prime      

9i 

of 

72 

Figure,  Plane      

123 

"                   "          Reduction  of. 

65 

Figures        

a 

"                  "         Simple    . 

61 

Fixed  pulley       

183 

"                  "         Subtraction  of 

70 

Follower      

Denominator       

23 

Foot-pound          

X9K 

"           Least  common      . 

27 

Force    .        

156 

Diagonal  of  parallelogram 

129 

"     Centrifugal       .... 

'94 

Diameter  of  circle      .... 

133 

"     Centripetal       .... 

195 

Difference    

9 

"     Direction  of      .... 

156 

"          (in  percentage) 

56 

"     Point  of  application  of    .        . 

156 

Digits  

2 

"     Pump         

343 

Direct  proportion  

101 

"     Representation  of    . 

H',,, 

"       ratio         

97 

"     required  to  punch  plates 

067 

Direction  of  force       .... 

156 

Formula      

us 

Dividend      .        . 

'7 

"        How  applied 

ixa 

Divisibility          

'Si 

"         Symbols  used  in 

"5 

Division       

i? 

Fraction       

•3 

"        of  decimals. 

45 

"        Improper      ,        .        .        . 

-5 

"         "  denominate  numbers    . 

73 

"        Proper           .... 

•t 

"         "  fractions. 

34 

"        Terms  of       .... 

»5 

Proof  of        .... 

20 

"       To  in  vert  a    . 

35 

"        Rule  for 

20 

"         "  reduce  a  decimal  to  a     . 

S* 

"        Sign  of         .... 

i? 

"         "        "      "  whole  or  mixed 

Divisor.        .        .        .        . 

'7 

number  to  a  . 

06 

Double  belts        

201 

"         "        "      to  a  decimal 

SP 

"        shear      

266 

"         "        "      "  lowest  terms  . 

>6 

Double-acting  pump 

245 

"          Value  of  a          ... 

94 

Downward  pressure  of  water  . 

209 

Fractions,  Addition  of 

•.•'i 

Driver  (pulleys)  

'73 

"          Division  of        ... 

34 

Dry  measure       .        . 

64 

"          Multiplication  of      . 

)a 

Ductility              

'S3 

"          Reduction  of    . 

25 

Duplex  steam  pump  .... 

246 

"          Roots  of    . 

94 

"          Subtraction  of 

3° 

E.                               PAGE 

"          To  find  least  common  de- 

Efficiency of  machine 

192 

nominator  of 

»7 

Elastic  limit        

248 

"           "  reduce  to  common  de- 

Elasticity           152, 

247 

nominator 

iS 

"          Measure  of        ... 

248 

Friction        

i8g 

Energy         

'99 

"         between    cross-head    and 

"        Conservation  of   . 

200 

guides       .... 

193 

Kinetic  

109 

"         Coefficient  of       ... 

Bg 

"        Potential       .... 

200 

Laws  of       .... 

90 

Epicycloidal  teeth      .... 

179 

"         Table  of  coefficients  . 

99 

Equality,  Sign  of                                 . 

4 

Frustum  of  cone  or  pyramid    . 

43 

Equilateral  triangle  .... 

126 

Altitude  of  . 

-i  i 

Evolution    

79 

"        Convex  surface  of 

44. 

Exhaustion  of  air       .... 

233 

"        Volume  of   . 

ft 

INDEX. 


PAGE 

PAGE 

Fulcrum      .... 

.     165 

Involute  teeth    .... 

.     179 

Fundamental  principle  of  machines    166 

Involution  

.      76 

G. 

PAGE 

Isosceles  triangle 

.    126 

Gain  or  loss  per  cent. 

.        60 

K. 

PAGE 

Gallon,  Cubic  inches  in 

.        65 

Kinetic  energy   .... 

.      199 

Gallon  of  water.  Weight  of 

.        65 

Gas,  Permanent 

.      150 

L. 

PAGE 

"    Tension  of  . 

222,  228 

Lacing  belts        .... 

.      203 

Gaseous  body 

.     mo 

Lateral  pressure  of  water 

.       212 

Gases,    Application    of    Mariotte's 

Law  of  Mariotte 

.       229 

law  to        .... 

•       230 

"     "  Pascal     . 

.       209 

Cear  wheel's 

Laws  of  friction. 

Gears,  Bevel 

•     '75 
•     !75 

"      "  inertia   .... 

•       157 

"        Horsepower  of 

204 

"      "  liquid  pressure    . 

.       209 

Miter 

•     J7S 

"     "  motion,  Newton's 

•        '57 

"           T?     1         f 

Lever 

ific 

"        Spur        .        ... 

178,  180 
175 

"      Compound 

105 

.   i67 

Gravity,  Center  of     .        .  • 

.     160 

"      Fulcrum  of     . 

.  i65 

"         Specific 

.     196 

"      Law  of     . 

.  i6S 

Guides,  Friction  on    . 

•     *93 

Lifting  pump      .... 

.  242 

Like  number       .... 

i 

H. 

PAGE 

Line,  Curved       .... 

.  119 

Hardness     .... 

.     152 

"      Horizontal 

.   119 

Head  (of  water)  . 

.     215 

"      Perpendicular  to  another 

.  119 

Helix    • 

.          .     187 

"      shafting     .... 

.  268 

Hemp  ropes         .        .        . 

.     252 

"           "         Bearings  for 

.  269 

"         "      Strength  of    . 

.     252 

"           "         Horsepower  of   . 

.  270 

Heptagon    .        . 

.      130 

"           "         Rules  for     . 

.  271 

Hero's  fountain. 

•     237 

"      Straight    .    •     . 

.  119 

Hexagon 

"      Vertical     .... 

Horizontal  line  . 

.     119 

Linear  measure. 

.    62 

Horsepower        .        .        • 

198 

Lines  Parallel 

I  TO 

."            of  belts. 

.        202. 

Liquid  body        .... 

.        150 

"             "  gears. 

.        204 

"       measure 

•           64 

"             "  shafting,    Fori 

nulas 

pressure,  Laws  of. 

.        20g 

for    . 

.    270 

Liquids,  Incompressibility  of  . 

.        207 

Hydraulic  press. 

.     216 

Local  value  of  a  figure      . 

2 

Hydrostatics 

.    207 

Long  ton  table    .        . 

•          63 

Hypotenuse 

.     126 

Loss  per  cent  - 

60 

I. 

PAGE 

M. 

PAGE 

Impenetrability. 

.      151 

Machines,  Pneumatic 

.      231 

Improper  fraction 

•        25 

"           Principle  of 

.     166 

Inclined  plane    . 

.      185 

Simple 

.     165 

"             "     Rule  for 

.     186 

Magdeburg  hemispheres  . 

•     233 

Incompressibility  of  liquids 

.     207 

Magnitude  of  a  force 

•      T57 

Indestructibility,  Law  of 

.     152 

Malleability         . 

•      153 

Index  of  a  root  . 

79 

Mariotte's  law    .        .        .        . 

.      229 

Inertia  , 

•       151.  158 

Materials,  Crushing  strength  of 

•      255 

Injector       .        .        .  .  ••  . 

•     239. 

"           Shearing         "          " 

.      266 

"        How  operated 

.     240 

"          Tensile            " 

.      248 

Inscribed  angle  . 

•      '33 

"          Transverse    "         " 

.     261 

polygon     . 

•      *34 

Matter  

•      M9 

Integer         .        .        . 

"       Properties  of  . 

•      IS' 

Inverse  proportion    . 

101,  104 

"      Special  properties  of     . 

•      152 

"       ratio 

97 

Means  (of  a  proportion)     . 

.        101 

PAGE 

O.                               PAGE 

Measure,  Cubic  . 

.       6, 

Obtuse  angle 

lao 

Dry      . 

.       64 

"          Linear          .        . 

62 

P.                          P 

AGE 

"          Liquid 

.       64 

Parallel  lines       

119 

"          of  money     . 

.       65 

Parallelogram    

i«3 

"           "  time 

64 

Altitude  of 

133 

"          Square 

62 

"             Area  of       ... 

124 

Measures,  Miscellaneous  . 

.       65 

"             Center  of  Gravity  of 

162 

of  angles  and  arcs 

.       64 

"             Diagonal  of       ... 

i*9 

"           "  capacity 

64 

Parallelopipedon        .... 

138 

"           "  extension 

62 

Parenthesis  53 

,  116 

"          "  weight  . 

63 

Partial  vacuum  . 

Mechanics    .... 

•      T49 

Pascal's  law        .                  ... 

Mensuration 

.      119 

Path  of  a  body  in  motion. 

'    1 

Minuend       .... 

9 

Percent  

'  I  ' 
53 

Minus   

9 

"      "     Gain     

Miscellaneous  measures    . 

65 

"      "     Loss 

6O 

Miter  gears. 

•     175 

"      "     Sign  of         .... 

55 

Mixed  number    .         .     -    . 

25 

Percentage  55,  56 

Mobility       .... 

•     151 

Rules  of. 

56 

Molecule      .... 

•     M9 

Perimeter  of  polygon 

Money,  Measure  of    . 

.         -       65 

Permanent  gas   . 

150 

United  States 

.         .       65 

Perpendicular  lines   .... 

"9 

Motion          .... 

•     '53 

Pillars,  Constants  for        .        .      257 

"      Newton's  laws  of   . 

•     '57 

"       Formulas  for  strength  of    . 

256 

Movable  pulley  . 

.     183 

Pinion  .        . 

J73 

Multiplicand 

Pitch  circle  

'77 

Multiplication 

12 

"     of  gear  teeth     .... 

177 

"              of  decimals 

43 

"      "  screw  thread 

187 

"               "  denominate 

num- 

Plane  figure         

"3 

bers   . 

72 

"         "       Center  of  gravity  of    . 

i6a 

"              "  fractions 

•      32 

"    Inclined     

95 

"              Proof  of 

.      16 

"        Law  of        .        . 

i   • 

Rule  for      . 

.       16 

Plunger  pump     ... 

»43 

Sign  of 

12 

Pneumatic  machines. 

931 

table  . 

13 

Pneumatics         

Multiplier    .        .        .        f 

12 

Point  of  application  of  force     . 

,.'. 

N. 

PAGE 

Polygon        

1    - 

Naught         .... 

"         Area  of         .... 

IOJ 

Newton's  laws  of  motion. 

•      157 

Inscribed     .... 

134 

Notation 

•     I.  4 

"         Perimeter  of        ... 

130 

"         Arabic 

2 

"         Regular        .... 

130 

Number 

"         Sum  of  interior  angles  of 

131 

"       Abstract 

I 

Porosity       

'  Jl 

"        Composite    . 

21 

Potential  energy 

• 

"       Concrete 

1 

Power  arm  

.65 

"       Denominate 

.      61 

"        of  a  number    .... 

;' 

"        Like       .        .        . 

i 

"        "  "  ratio          .... 

99 

"        Mixed    ... 

25 

Powers  and  roots  in  proportion 

"       Prime    .        .      -. 

21 

Press,  Hydraulic 

IT« 

"       Reciprocal  of  a    . 

•       97 

Pressure  of  air    

.•-•; 

Unlike                   ,     • 

"          "  atmosphere    . 

,.•; 

"        Unit  of  a 

I 

"  liquid       .... 

g 

Numeration 

.     I,  4 

"          "  water       .... 

PAGE 

R.                               PAGE 

Pressure  of  water,  Lateral 

212 

Radical  sign        

n 

'     Upward 

211 

Radius  of  circle  

i  a 

Prime  factor        

21 

Rate  (in  percentage)  .... 

56 

"      number    

21 

Ratio    .        

96 

Principle  of  Archimedes  . 

220 

"     Couplet  of  a 

•a 

"          "  machines 

1  66 

"     Direct         

91 

Prism   

138 

"•     How  expressed 

9« 

Convex  surface  of 

'39 

"     Inverse       

97 

"       Volume  of 

'39 

"      Reciprocal         .... 

'~<7 

Product        '        

12 

Symbol  of          .... 

<-/> 

Partial  ..... 

15 

"     Terms  of  a         .... 

V 

Proof  of  addition        .... 

8 

"     Value  of  a          .... 

91 

"      "  division         .... 

20 

"     Velocity     

189 

"      "  multiplication 

:6 

Reading  numbers       .... 

3 

"      "  subtraction  .... 

n 

Reciprocal  of  a  number    . 

«7 

Proper  fraction  

25 

"  "  ratio 

91 

Properties  of  matter  .... 

IS' 

Rectangle    

1*3 

Proportion  

100 

Reduction  of  decimals    to  fractions 

51 

"           Compound 

109 

"           "    denominate  numbers 

•.-, 

"           Couplet  of        ... 

100 

"           "   fractions 

25 

"           Direct       .... 

101 

"           "          "          to  decimals 

SO 

"           Extremes  of     . 

101 

Regular  polygon         .... 

130 

"           how  read. 

100 

Remainder  

9 

"           Inverse      .        .        .      101, 

104 

Rhomboid    

123 

"           Means  of  a 

101 

Rhombus     

1*3 

Powers  and  roots  in 

106 

Right  angle          

120 

Rules  for           ... 

101 

Right-angled  triangle 

u6 

"           Simple       .... 

109 

Ring,  Circular     

1*6 

Pulley,  Fixed      

183 

"             "         Area  of       ... 

i;'' 

Movable         ....        . 

'83 

"      Cylindrical,  Area  of 

146 

Pulleys         

170 

"                "            Volume  of  . 

14? 

"       Balanced       .        .        .  .  -     . 

171 

Root      

77 

"       Combination  of    .        .  "    . 

184 

"      Cube.         .                  .         .         79 

86 

"       Crowning  of. 

171 

"     Index  of    

19 

"       Law  of  

.85 

"     of  gear  tooth     .... 

'77 

"       Rules  for       .... 

172 

"     "  ratio        

I'M 

Pump,  Air   

23' 

'•     Square       

79 

"        "    chamber  for 

244 

Roots  of  fractions       .        .        .     ^  . 

94 

"      Force       

242 

11     other  than  square  and  cube  . 

95 

"       Lifting     

242 

Rope,  Hemp 

j:j 

"       Plunger   . 

243 

"     Strength  of   . 

t$9 

Steam      

245 

"       Wire        .... 

•S3 

Suction    

241 

"     Strength  of   .        .... 

•S3 

Punching  holes  in  plates,  Force  re- 

quired for 

267 

S.                                  PA 

GE 

i  'fj 

Altitude  of  .        .        .        . 

142 

Screw  

"         Convex  area  of  . 

142 

"      Law  of               .        ... 

t88 

**          Frustum  of 

*v      Pitch  of 

,x- 

"         Slant  height  of    . 

•43 

142 

"      thread       .        .        .     '   .      '  . 

««7 

Vertex  of     ... 

142 

Sector,  Area  of   

«3I 

"         Volume  of   . 

142 

Segment,  Area  of 

«37 

Semicircle    

":4 

Q.                                  PAGE 

Semi-circumference  .... 

i  ;4 

Quadrilateral      

123 

Shafting,  Black  .... 

•M 

Quotient       

17 

"         Bright         .... 

tU 

INDEX. 


PAGE 

PAGE 

Shafting,  Cold-rolled 

268 

Subtraction 

9 

"         Distance    between    bear- 

«   of  decimals    . 

42 

ings  of     .... 

269 

"            "  denominate    num- 

"        Horsepower  of  . 

270 

bers 

7° 

"         Line     

268 

"             "  fractions   . 

3° 

"         Rules  for    .... 

271 

Proof  of 

ii 

Shear,  Double    

266 

Rule  for         ... 

ii 

"      Single       

266 

Sign  of  . 

9 

Shearing  strength      .... 

266 

Subtrahend          

9 

Rule  for       .     '  . 

267 

Suction  pump    

241 

Table  of      . 

266 

Surface  of  solid  .... 

139 

Sign  of  addition          .... 

4 

Symbol  of  ratio  

96 

"     "    division          .... 

'7 

Symbols  of  aggregation    . 

53 

"     "    dollars  

53 

"      "    equality          .... 

4 

T.                         i 

'AGE 

•'     "    multiplication 

12 

Table,  Addition          .... 

5 

"     "    per  cent  

55 

"       Multiplication 

13 

"      "    subtraction   .... 

9 

"      of  coefficient  of  friction 

192 

"    Radical        

79 

"       "  constants  for  cast-iron  pil- 

Simple denominate  number     . 

61 

lars    . 

258 

"        machines        .... 

165 

Table  of  constants  for  transverse 

"       proportion      .... 

109 

strength  . 

261 

"        value  of  figure 

2 

"       "           "          "  wooden  pil- 

Single belts          ..... 

201 

lars   . 

259 

Siphon          

238 

"       "           "          "  wrought-iron 

Slant  height  of  cone  or  pyramid 

142 

pillars 

257 

Solid  body   

I38 

"       "  crushing  strength 

256 

"     Center  of  gravity  of 

l64 

"       "  distances    between   shaft 

Specific  gravity          .... 

I96 

bearings  .... 

269 

Sphere          

M5 

"       "  shearing  strength 

266 

"      Area  of  surface  of 

145 

"       "  tensile  strength 

249 

"      Volume    . 

146 

Teeth  of  gears    

177 

Spur  gears  

175 

"       "       "     Addendum  of  . 

'77 

"         "     Horsepower  of 

204 

"       "       "     Epicycloidal 

179 

Square          

ir>3 

"       "       "     Involute    . 

179 

"       foot  .        .        .        .        . 

124 

"       "       "     Pitch  of     ... 

'77 

"       measure  

62 

Tenacity      

'77 
'53 

"       of  a  number    .... 

77 

Tensile  strength         .... 

248 

"       root  

79 

"               "         of  chains 

251 

"           "    Rule  for          ... 

84 

"               "         "  materials  . 

249 

"           "    Proof  of  . 

84 

"               "         "  ropes 

252 

"           "    Short  method  for 

85 

Rules  for       . 

250 

Steam  pump        

245 

Table  of 

249 

Strain  

247 

Tension  of  gases         .        ._       .222 

,  228 

Strength,  Compressive 

255 

Terms,  Higher  

25 

"         of  beams    .... 

262 

"        Lower    

26 

"          "  belts        .... 

202 

"        of  a  fraction  .... 

25 

"          "  chains     .... 

2S' 

"         "  "  proportion 

101 

"          "  materials 

247 

"         "  "  ratio         .... 

97 

"          "  pillars,  Formulas  for  . 

256 

Thread  of  screw         .... 

187 

"          "  ropes  (hemp  and  wire' 

252 

Time,  measures  of     .... 

64 

"         Shearing    .... 

266 

Torricellian  vacuum 

234 

"         Tensile        .... 

248 

Train    

'73 

Stress   

247 

Transverse  strength 

261 

"      Unit  

247 

"                   "         of  beams 

261 

Transverse  strength  of  beams, 

PAGE 

Vertex  of  angle  .... 

PAGE 

.        120 

Rules  fo 

r      262 

"•        '•  cone    .... 

•      '44 

"         "  cantilevers     263 

Vertical  line        .... 

.     nq 

"                 "         "  constants 

.    261 

Vinculum    

S3.  1.6 

Trapezoid    ... 

•    "3 

Volume  of  circular  ring    . 

.      147 

"          Area  of              .        . 

.    124 

"        "  cone  .... 

.      142 

Triangle      

,    126 

"        "  cylinder   . 

•      139 

Altitude  of 

.   129 

"        "  cylindrical  ring 

•      147 

"         Center  of  gravity  of 

.    162 

"        "         •'        "  pyramid 

742 

"         Equilateral 

.    126 

•'        "  prism 

.      139 

"         Hypotenuse  of    . 

.    126 

•'        "  pyramid    . 

.      142 

Isosceles 

.    126 

"        "  sphere 

.     146 

"         Right-angled 

.    126 

Unit  of  .        . 

•      139 

"         Scalene 

.    126 

Troy  weight       .... 

•    63 

W. 

PAGE 

Water,  Buoyant  effect  of 

.      219 

U. 

PAGE 

"       Gallons  in  a  cubic  foot  of 

65 

Uniform  velocity 

•      '54 

"       Incompressibility  of 

.      207 

Unit      

I 

"       Pressure  of     .        . 

.      209 

"    of  a  number 

I 

Wedge          

.      187 

"    square          .... 

.      124 

Weight         

•      'SI 

United  States  money 

.        65 

"        Arm        .... 

.      165 

Unlike  number 

I 

"        Avoirdupois  . 

•        63 

P            P 

"        Measures  of  . 

•      234 

•        63 

V. 

PAGE 

Troy        .... 

.        63 

Vacuum       

.      224 

Wheel  and  axle  .... 

.      169 

Partial 

,      224 

Wheelwork          .... 

•      173 

"         Torricellian 

•      234 

Wheelwork,  Rules  for 

•      '74 

Value  of  a  fraction     .        .     .  .• 

24 

Wire  rope,  Strength  of 

•     253 

"    ratio 

•       97 

Work    

.     197 

Vapor  

•       5° 

"      Measure  of        ... 

•      197 

Variable  velocity 

•       54 

"      Unit  of       .... 

.      197 

Velocity       

54 

Worm  and  worm-wheel     . 

.      176 

"         ratio      .... 

.       89 

"         Uniform 

54 

Z. 

PAGE 

variable 

54 

Zero       * 

2 

5?S>3  DZ>  >I>  D  >^SDO  3^.d 


. 


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O  ^?»DD> 


^ 


CCX 


AA    000513599 


wccggx. 

CSCCCJK 
mOTOf 

recede 

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